Mai jab bhi lecture kholne baithta hu to gharwale piche padd jate hai ki school me nahi padhaya kya......kabhi nahi. samajhte...unko lagta hai ki school ki Copy ke question ans ratne se board me top maar dunga.....pareshan ho chuka hu
[Some important and extra points] 7:50 (intensity of light) 10:40 (what happens when light incident) 13:30 (How do we see) 17:15 (bounces back into the same medium) 18:00 (normal is drawn to find
Summary: Light can be viewed as both a particle (photon) and a wave. The concepts of reflection, refraction, and the properties of lenses are discussed in detail. Key moments: 00:00 Light has dual nature, exhibiting both particle-like and wave-like properties. The intensity of light depends on the number of photons, with brighter light having more photons, showing its particle nature. -Historical evolution of the understanding of light from Newton's corpuscular theory to Einstein's photon theory. -Discussion on the electromagnetic wave nature of light and how it can be perceived by the human eye compared to other wavelengths in the electromagnetic spectrum. -Exploration of the intensity of light based on the number of photons, highlighting the relationship between brightness and the quantity of photons. 10:05 Light rays reflect off surfaces because they need a source of light to be visible. Reflection occurs when light bounces off a surface, following the laws of incident and reflected angles. -Explaining the importance of a light source for visibility and the role of reflection in making objects visible. -Understanding the behavior of light rays when they hit surfaces and the concept of reflection. -Discussing the laws of incident and reflected angles in reflection, emphasizing their equal importance in determining the direction of reflected light. 20:07 Understanding the concept of plane mirrors involves incident rays, reflected rays, and normal lines, leading to the creation of images with specific properties and characteristics. -Exploring the concept of plane mirrors and incident rays. It involves understanding reflections and angles of incidence. -Discussing the nature of images created by plane mirrors. It includes the distance, height, and characteristics of the reflected image. -Explaining lateral inversion in plane mirrors. It details how images appear laterally inverted and the concept of virtual images. 31:12 Understanding the concepts of concave and convex mirrors is crucial in optics. Concave mirrors converge light rays, creating real images, while convex mirrors diverge light rays, forming virtual images. -Differentiating between concave and convex mirrors based on their light convergence or divergence properties is essential for image formation. -Real images are produced by concave mirrors when light rays converge, while virtual images are formed by convex mirrors when light rays diverge. 40:29 Understanding the concepts of spherical mirrors and image formation involves terms like center of curvature, principal axis, and focal length. These concepts help determine the nature and size of the image formed by the mirror. -Explanation of the terms related to spherical mirrors like center of curvature and principal axis. These terms play a crucial role in understanding mirror properties. -Discussion on how the position of the object relative to the mirror affects the nature and size of the image formed. Concepts of real and inverted images are explained. -Importance of placing the object correctly in relation to the mirror to determine the characteristics of the image formed. The role of object position in image formation is highlighted. 56:00 Understanding the nature of images formed by convex mirrors is crucial. Convex mirrors always create virtual, erect, and diminished images, making objects appear smaller and creating a specific field of view. -Characteristics of images formed by convex mirrors. They are always virtual, erect, and diminished in size, providing a unique perspective of the object being reflected. -Field of view created by convex mirrors. Convex mirrors offer a larger field of view, allowing for a broader perspective and making objects appear smaller and farther away. 1:01:06 The video explains the concepts of Cartesian planes, sign conventions, and mirror formulas. It emphasizes understanding the origin, positive and negative values, and magnification in mirror images. -Explanation of Cartesian planes and sign conventions. -Detailed explanation of mirror formulas and magnification in mirror images. -Importance of practice and hard work in understanding and applying the concepts. 1:11:14 To achieve big goals in life, one must step out of their comfort zone and challenge themselves. Overcoming comfort zones is essential for significant achievements and personal growth. -Importance of pushing boundaries for personal growth and success. Stepping out of comfort zones leads to self-improvement and achieving big goals. -Explanation of mirror formulas and numerical problems in optics. Understanding focal lengths, image positions, and mirror types in physics. -Different types of mirrors and their characteristics. Exploring concave and convex mirrors, their focal points, and applications in optics. 1:21:23 The video discusses the concept of refraction of light, explaining how light changes direction when passing through different transparent mediums with varying optical densities. -Explanation of different transparent mediums and their refractive indices like vacuum, glass, water, kerosene, and diamond. -Understanding the relationship between optical density, refractive index, and the behavior of light when transitioning between mediums of different densities. -Introduction to the rules of light transition between different mediums, including the behavior of light rays in relation to the normal line and angles of incidence. 1:31:32 Understanding the concept of refraction when light falls normally, leading to no refraction, and discussing lateral displacement and spherical lenses in optics. -Explaining the concept of normal incidence and zero refraction when light falls perpendicular to a surface. -Discussing the absence of refraction and the behavior of light in different mediums. -Introducing the concept of lateral displacement and moving on to understanding spherical lenses in optics. 1:41:36 Understanding the rules of focal length and image formation in lenses is crucial for determining the nature and size of images. Sign conventions play a key role in simplifying numerical calculations in optics. -Explanation of focal length and image formation rules in lenses, emphasizing the impact on the nature and size of images. -Sign conventions in optics and their significance in numerical calculations for determining object positions and image characteristics. 1:53:50 Understanding the lens formula and focal length calculations are crucial in optics. The video discusses how to determine object distance, image distance, and focal length using the lens formula and specific measurements. -Explanation of lens formula and focal length calculations for optics applications. It emphasizes the importance of understanding object distance, image distance, and focal length. -Discussion on the nature of lenses, including convex and concave lenses, and their impact on image formation. Exploring the concept of magnification and the significance of lens types in optics. -Introduction to the concept of power in lenses and the calculation of power using diopters. Understanding the relationship between power and focal length in optics applications. 2:07:02 Refractive index explains how light changes speed in different mediums. Absolute and relative refractive indices differ in how they consider the medium. The formula for refractive index relates to Snell's Law and constant frequency. -Difference between absolute and relative refractive indices. Absolute considers light coming from another medium, while relative considers the order of the mediums. -Explanation of refractive index and how it relates to the speed of light in different mediums. The formula involves the ratio of speeds in different mediums. -Relationship between refractive index formula and Snell's Law. The constant frequency in the formula ensures the speed of light remains consistent. 2:14:16 The video discusses the concept of refractive index and its calculation, emphasizing the importance of understanding the formula and shortcuts for solving related problems effectively. -Refractive index explained through formulas and concepts. Understanding the relationship between frequency, medium, and wavelength. -Absolute refractive index and its significance in calculations. Exploring the formula for refractive index and its application. -Utilizing shortcuts for calculating refractive index efficiently. Demonstrating how to simplify calculations using shortcuts and practical examples. kam ayega likh lo
1:19:31 answer is v = 3.75cm m = 0.375 HEIGHT OF IMAGE = 0.75 NATURE: virtual , erect , diminished . 1:21:10 ANSWER IS V = 60/7CM M=3/7 Height of image = 15/7 NATURE: virtual , erect , diminished. (THANK YOU SIR FOR THE GREAT VIDEO)
@@Sunny-yj9gt I think, v = 15/4 and in decimal form it is 3.75 You are asking about this question - An object is 2cm tall is placed at 10 from a diverging mirror of focal length 6cm . Find the nature and position of the image
Bhai aapne galat kardiya hai, v ko simplify nahi karna tha , usse end me m ki calculation ke dauran as it is use karna tha Taaki Phir Baad me height .75 aaye aur m = .375 ho
1:09:55 Sir this case is of concave mirror and it is case no 4 where the object is placed between C and F and the image formed beyond C nature is real,inverted and enlarged
As it is diverging mirror, so it is convex mirror. Given→ Height of Object= 2 cm u= -10 cm f= 6 cm Solution→ 1/v + 1/u = 1/f 1/v + 1/(-10) = 1/6 1/v - 1/10 = 1/6 1/v = 1/6 + 1/10 1/v = (5+3)/30 = 8/30 = 4/15 v = 15/4 cm m = -v/u = -(15/4)/-10 m = (15/4)/10 m = (15)/(4*10) m = 15/40 m = 3/8 = 0.375 times m = Hi/Ho 3/8 = Hi/2 (3/8)*2 = Hi Hi= 3/4 = 0.75 cm Nature of image→ As, m is positive so, the image is virtual and erect, and As, m
2:22:16 Sir isme humlog 42 na leke image aur object ka distance to diya hi hua hai to usi se nikal sakte hai na 42 lekar hi kyu nikale? Pls help me sir
Numericals on concave and convex mirror 1:14:09 - object at C 1:16:46 - object at infinity Homework question 1:19:37 - v= 15/4 m= 3/8 Erect,virtual and diminished Object at infinity 1:21:08 - f= 15 cm v= 60/7 m= 3/7 Erect, virtual and diminished Object at infinity
Digraj sir aap bhut hi acha padhate ho Isi liye maine 11me humanities opt ki hai Mere s.s.c pre board se phle 2din ka gape tha . Usse phle mene history ki book open bhi nhi ki thi Bss aap ke lecture dekhe the history ke or mere s.s.c me 78/80 THANK YOU SO MUCH SIR
00:02 Understanding the nature of light and its properties 02:08 Light is not considered a particle but rather an electromagnetic wave. 06:22 Light is an electromagnetic radiation that is visible to us. 08:28 Light travels in straight paths 12:46 Light is absorbed, transmitted, and reflected by surfaces. 15:09 Light reflection enables us to see objects 20:06 Angle of incidence and angle of reflection are always equal 22:55 Understanding reflection and refraction of light. 27:03 Understanding real and virtual images 28:55 Understanding reflection and refraction angles 34:44 Understanding focal length and radius of curvature 37:03 Understanding light reflection and refraction 41:39 Understanding rules for creating the image 45:07 Reflection and refraction of light explained through the concept of infinity in physics 51:57 Understanding the properties of convex mirrors 56:49 Understanding reflections and refractions in concave and convex mirrors 1:01:14 Understanding the sign convention of reflection and refraction on a Cartesian plane 1:03:09 Focal length of concave and convex mirrors 1:08:33 Inverted image signifies vertical formation 1:10:36 Understanding the importance of hard work and getting out of the comfort zone 1:14:56 Understanding the nature and position of M for reflection and refraction 1:18:59 Light reflection and refraction key points 1:23:56 Refraction caused by difference in optical density 1:25:48 Refraction and reflection of light 1:32:14 Understanding the concept of refraction and reflection. 1:35:11 Light undergoes lateral displacement and changes in wavelength and angle in glass 1:41:38 Understanding focal length and its effect on light 1:45:00 Understanding image formation in reflection 1:53:46 Understanding light reflection and refraction in concave and convex lenses 1:55:29 Understanding lens formula and image formation. 1:59:56 Understanding object distance and image formation in reflection and refraction. 2:01:59 Understanding power of a lens and its formula 2:06:47 Understanding refractive indices in reflection & refraction 2:08:48 Understanding refractive index and its relation to speed of light 2:12:50 Refraction of light and refractive index explained 2:15:20 Understanding absolute and relative refractive index 2:21:26 Solving for nx and discussing the answers
The lecture is about 2 hrs 22 mins. But it actually takes about 2 days to understand everything. To practice between the lectures... Thank you so much sir for helping us this much for free.
29:58 The angle R2 will be 30 degrees 1:09:55 It is the case of concave mirror when the object is kept in between c and f 1:21:08 V=-10cm Nature of image is Real,Inverted and diminished 1:56:53 V=-30/7cm Nature of image is Real,Inverted and diminished 1:57:29 v=15cm,hi=3cm Natur of image is virual,erect and enlarge
FROM : PRASHANT KUMAR 1:19:31 answer is v = 3.75cm m = 0.375 HEIGHT OF IMAGE = 0.75 NATURE: virtual , erect , diminished . 1:21:10 ANSWER IS V = 60/7CM M=3/7 Height of image = 15/7 NATURE: virtual , erect , diminished. (THANK YOU SIR FOR THE GREAT VIDEO )👍👍
ok see now, sol:- u= -10 ho= +2 f = +6 1/f = 1/v + 1/u so calulation ans. will 8/30=1/v simplify it then 4/15 = 1/v then v= 15/4 now, m=-v/u -15/4\-10/1 solve it ans will m= -3/8 @@killerxgaming2491 sol
@@Ankit.Raj_2509Bro actually @Ayush123m is wrong because he forgot to solve the hô M = hi/ho = -3.75 (-v)/-10 (u) then - cancelled and then 3.75 will be 0.375 the ho = 2 will be transpose to RHS and then 0.375×2 = 0.75 and this is the complete solution! Virtual+Erect+Diminished
1:09:46 - Case 4 of concave mirror i.e The object will be placed b/w C and F. 1:19:43 - hi= +0.75cm ? 1:25:15 height of image +2.14 and nature virtual , errect and diminished for both questions
1:19:37 (v=3.75cm,m=0.375cm,hi=0.75cm)Nature of image is virtual and erect, position of image is behind the mirror, size of image is diminished (smaller than the object).
After knowing the value of v we can directly write the size of images as it is form in convex mirror and we know that convex mirror always make diminished image?
27 OCTOBER 24! LIGHT DONE { js need to do some more numericals} 1:19:42 v= 15/4 cm , m = 3/8 , hi=+0.75cm. The images virtual erect and diminished. 1:22:14 v=60/7 cm , m = 3/7 , hi =15/7 cm. 1:56:47 v=-30/7 cm , 2/7=m. 1:57:02 v=15/2cm m= -3/4 and -3/2cm. ( i may be wrong somewhere or forgor to add signs ;-; ) WELZZ, TY SIR
Rakshak sir never found a teacher so pure and helpful like you thank you sir.....Half yearly me bhi aapki wajah se 24/26 aaye or pre board me bhi aapki wajah se marks aaye aab boards phodungi .....thank you sir love from Udaan 2.0 Lots of respect from Kabutar🥺❤️🔥 I got 95 out of 100 in Science in my boards and I literally watched this lecture one night before my board exam. PW is emotion❤️
Q 14 Answer :- Given:- Ho =+5 cm u = -20cm f = +15cm Solution : Using mirror formulae 1/f = 1/v +1/u 1/15 = 1/v - 1/20 1/20 + 1/15 = 1/v After taking LCM 15+20/300 = 1/v 7/60 = 1/v v = 60/7 Now, m = -v/u m = -60 - 7 - -20 - 1 m = 3/7 m = hi/ho 3/7 = hi/5 15/7 = hi hi = 2.14 cm Therefore, height of image (hi) = 2.14cm and nature of image is virtual & errect & diminished, because (m>1) Thankyou... ❤
@@X_euphoriax..x given, ho =2cm u = -10cm f = 6cm (convex mirror has focus on right side) We know, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/6 - (1/-10) = 1/6 + 1/10 = 5+3/30 = 8/30 Therefore, v = 30/8 = 3.75cm We know, m= -v/u = -3.75/-10 = 0.375 (We get to know that the position is virtual, erect and diminished) To find the size of image, 0.375 = hi/2 ---( m = Height of image/Height of object) hi = 0.75cm. Hope this helped you!
29:56 30° 01:09:55 4) Object between centre of curvature and principal focus. 01:19:44 m = +3/8 , v = 15/4 , height of image = 3/4. 01:21:10 v = 60/7, height of image = 15/7, m = +3/7. 01:56:54 v = 15 , m = -3/2 , height of image = -3. A very heartly thanks to rakshak sir for all this❤
1:56:53 answer will be v=-30/7, m=2/7 that means nature of the image will be diminished, virtual,erect.. position of the image will be at between f1 and O
1:09:53 Ans CASE 3: when object is placed between Center of curvature and focus the image will form beyong C Characteristics : • Real and inverted • Beyond C • Enlarged • Concave Mirror
1:56:47 v=-4.3 cm and m=0.28 Hence image will be virtual erect and diminished 1:57:29 v=15cm height of image will be 3 and nature of image will be virtual erect and enlarged
@@triadictable9979 bhai 'Hi' ya m ke value pe + ya - se pata chalta hai V pe negative positive ye pata ni chlta V pe + ya - to position bata ta h image ka naa ki nature
@@Siddharth_10_ bro,you must watch the video of HC Verma regarding this topic. He is the best physics professor in india. He said that bending is only indication of refraction,but refraction is itself not bending
@@Siddharth_10_ yes bro I agree that, and that is why ncert had not touched the topic of normal incidence. Because the writers of ncert know that it would become a controversial topic.Even though in some books this topic is mentioned,but there it is written that in case of normal incidence the light goes undeviated. It isn't mentioned that refraction does not takes place.
1:19:36 ans = distance of image or v = 15/4cm. And magnification of image = 3/8. height of image = 15/2 cm and nature of image = virtual erect and diminished as its a convex mirror
Sir in the first question 30:57 we can also asuume rays as parallel and apply co interior angles property? Because the answer will be 30 only as we know that both angles are equal.
Bhai मैं state board wala hu lekin padhai ki chul ke karan mai ise dekh raha hu kyuki mene ek mahine me school ka to syllabus khatam kr diya ab cbse ki bari
1:33:13 Sir refraction occurs in the case of normal incidence but there is no bending of light. Because Refraction also means the change in speed of light.
1:19:36 My answer is -: 1. V=30/8 2. M=3/8 3. Height of object= 6/8 4. Nature -: Virtual+erect+diminished --- 2 case of convex mirror ( between P and F )
@@AyshaAli-wr6nzhi/ho=magnification And magnification hamne already calculate kiya tha which was equal to 3/8 Now-- hi/ho=3/8 Ho=2cm(given) Hi/2=3/8 8hi=6 Hi=6/8=3/4 I hope u understood
1:9:35 homework ques Height of object =2 distance= -10 position of image = 3.7cm height of image = 0.75cm nature of image = virtual and eract, diminished hope my and is correct
1:19:40 the nature of the image is virtual and erect, the position of the image is 3.75cm, between infinity and pole of the mirror, the size of the image is 0.75cm
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Mai jab bhi lecture kholne baithta hu to gharwale piche padd jate hai ki school me nahi padhaya kya......kabhi nahi. samajhte...unko lagta hai ki school ki Copy ke question ans ratne se board me top maar dunga.....pareshan ho chuka hu
ekdam sahi bola bhi mere ghar me bhi yahi hota hai tang aa gya hu😢😢 edit thanks for 14 likes🎉😢😢😢😢
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Same bro
Samjhao bhai gharwalo ko aur koi option bhi toh ni h 🤔
[Some important and extra points]
7:50 (intensity of light)
10:40 (what happens when light incident)
13:30 (How do we see)
17:15 (bounces back into the same medium)
18:00 (normal is drawn to find
This is important topics 👍👍
Hi didi 😊
Thank you ma 😊
Hii
heyyy!!!bsss aa😂
The one who can compresshed the 8hour's ch in 2 and half an hour with full concept is RAKSHAK SIR❤❤❤.
Sir ka 16 hrs ka total lecture hai lol
@@Yashuu-69 mmm🎉
@@Yashuu-69kaha hai
AND WITH PYQ
Bhai jitna kam time hoga utna kharab hai😢
Summary:
Light can be viewed as both a particle (photon) and a wave. The concepts of reflection, refraction, and the properties of lenses are discussed in detail.
Key moments:
00:00 Light has dual nature, exhibiting both particle-like and wave-like properties. The intensity of light depends on the number of photons, with brighter light having more photons, showing its particle nature.
-Historical evolution of the understanding of light from Newton's corpuscular theory to Einstein's photon theory.
-Discussion on the electromagnetic wave nature of light and how it can be perceived by the human eye compared to other wavelengths in the electromagnetic spectrum.
-Exploration of the intensity of light based on the number of photons, highlighting the relationship between brightness and the quantity of photons.
10:05 Light rays reflect off surfaces because they need a source of light to be visible. Reflection occurs when light bounces off a surface, following the laws of incident and reflected angles.
-Explaining the importance of a light source for visibility and the role of reflection in making objects visible.
-Understanding the behavior of light rays when they hit surfaces and the concept of reflection.
-Discussing the laws of incident and reflected angles in reflection, emphasizing their equal importance in determining the direction of reflected light.
20:07 Understanding the concept of plane mirrors involves incident rays, reflected rays, and normal lines, leading to the creation of images with specific properties and characteristics.
-Exploring the concept of plane mirrors and incident rays. It involves understanding reflections and angles of incidence.
-Discussing the nature of images created by plane mirrors. It includes the distance, height, and characteristics of the reflected image.
-Explaining lateral inversion in plane mirrors. It details how images appear laterally inverted and the concept of virtual images.
31:12 Understanding the concepts of concave and convex mirrors is crucial in optics. Concave mirrors converge light rays, creating real images, while convex mirrors diverge light rays, forming virtual images.
-Differentiating between concave and convex mirrors based on their light convergence or divergence properties is essential for image formation.
-Real images are produced by concave mirrors when light rays converge, while virtual images are formed by convex mirrors when light rays diverge.
40:29 Understanding the concepts of spherical mirrors and image formation involves terms like center of curvature, principal axis, and focal length. These concepts help determine the nature and size of the image formed by the mirror.
-Explanation of the terms related to spherical mirrors like center of curvature and principal axis. These terms play a crucial role in understanding mirror properties.
-Discussion on how the position of the object relative to the mirror affects the nature and size of the image formed. Concepts of real and inverted images are explained.
-Importance of placing the object correctly in relation to the mirror to determine the characteristics of the image formed. The role of object position in image formation is highlighted.
56:00 Understanding the nature of images formed by convex mirrors is crucial. Convex mirrors always create virtual, erect, and diminished images, making objects appear smaller and creating a specific field of view.
-Characteristics of images formed by convex mirrors. They are always virtual, erect, and diminished in size, providing a unique perspective of the object being reflected.
-Field of view created by convex mirrors. Convex mirrors offer a larger field of view, allowing for a broader perspective and making objects appear smaller and farther away.
1:01:06 The video explains the concepts of Cartesian planes, sign conventions, and mirror formulas. It emphasizes understanding the origin, positive and negative values, and magnification in mirror images.
-Explanation of Cartesian planes and sign conventions.
-Detailed explanation of mirror formulas and magnification in mirror images.
-Importance of practice and hard work in understanding and applying the concepts.
1:11:14 To achieve big goals in life, one must step out of their comfort zone and challenge themselves. Overcoming comfort zones is essential for significant achievements and personal growth.
-Importance of pushing boundaries for personal growth and success. Stepping out of comfort zones leads to self-improvement and achieving big goals.
-Explanation of mirror formulas and numerical problems in optics. Understanding focal lengths, image positions, and mirror types in physics.
-Different types of mirrors and their characteristics. Exploring concave and convex mirrors, their focal points, and applications in optics.
1:21:23 The video discusses the concept of refraction of light, explaining how light changes direction when passing through different transparent mediums with varying optical densities.
-Explanation of different transparent mediums and their refractive indices like vacuum, glass, water, kerosene, and diamond.
-Understanding the relationship between optical density, refractive index, and the behavior of light when transitioning between mediums of different densities.
-Introduction to the rules of light transition between different mediums, including the behavior of light rays in relation to the normal line and angles of incidence.
1:31:32 Understanding the concept of refraction when light falls normally, leading to no refraction, and discussing lateral displacement and spherical lenses in optics.
-Explaining the concept of normal incidence and zero refraction when light falls perpendicular to a surface.
-Discussing the absence of refraction and the behavior of light in different mediums.
-Introducing the concept of lateral displacement and moving on to understanding spherical lenses in optics.
1:41:36 Understanding the rules of focal length and image formation in lenses is crucial for determining the nature and size of images. Sign conventions play a key role in simplifying numerical calculations in optics.
-Explanation of focal length and image formation rules in lenses, emphasizing the impact on the nature and size of images.
-Sign conventions in optics and their significance in numerical calculations for determining object positions and image characteristics.
1:53:50 Understanding the lens formula and focal length calculations are crucial in optics. The video discusses how to determine object distance, image distance, and focal length using the lens formula and specific measurements.
-Explanation of lens formula and focal length calculations for optics applications. It emphasizes the importance of understanding object distance, image distance, and focal length.
-Discussion on the nature of lenses, including convex and concave lenses, and their impact on image formation. Exploring the concept of magnification and the significance of lens types in optics.
-Introduction to the concept of power in lenses and the calculation of power using diopters. Understanding the relationship between power and focal length in optics applications.
2:07:02 Refractive index explains how light changes speed in different mediums. Absolute and relative refractive indices differ in how they consider the medium. The formula for refractive index relates to Snell's Law and constant frequency.
-Difference between absolute and relative refractive indices. Absolute considers light coming from another medium, while relative considers the order of the mediums.
-Explanation of refractive index and how it relates to the speed of light in different mediums. The formula involves the ratio of speeds in different mediums.
-Relationship between refractive index formula and Snell's Law. The constant frequency in the formula ensures the speed of light remains consistent.
2:14:16 The video discusses the concept of refractive index and its calculation, emphasizing the importance of understanding the formula and shortcuts for solving related problems effectively.
-Refractive index explained through formulas and concepts. Understanding the relationship between frequency, medium, and wavelength.
-Absolute refractive index and its significance in calculations. Exploring the formula for refractive index and its application.
-Utilizing shortcuts for calculating refractive index efficiently. Demonstrating how to simplify calculations using shortcuts and practical examples.
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sider se nikali hai summary??
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Rakshak Sir Is really a rakshak Of 10 class students 🙂🙂
Yes I agree with your words
YEAHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
Of 9th students too
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New student in 10 class 2024-2025 batch..🎉
I am also
Me
@@manjukumari108 hiii manju
Yupp !!
@@rakshitabhati hiiiii
Trick to learn:
RIVER
RI- real and inverted
VE- virtual and erect
Aur jate jate like kar ke jana sab😅
Thank You Ms.Pragya
R ka kya hai behen
@@devitasharma09 nothing just ignore it
@@devitasharma09 ap ko Pata he to ap Bata dejiye
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Any students jo 2025 m board Dene vala h 😢
Same
mee
..
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Present 😅
1:19:31 answer is
v = 3.75cm
m = 0.375
HEIGHT OF IMAGE = 0.75
NATURE: virtual , erect , diminished .
1:21:10 ANSWER IS
V = 60/7CM
M=3/7
Height of image = 15/7
NATURE: virtual , erect , diminished.
(THANK YOU SIR FOR THE GREAT VIDEO)
Yes
Hey I have doubt m=3/2
Height of image= 15/2 =7.5
Is this right ?
In first ans V=10/3 , m = 1/3, h’= 2/3 is it?
@@Sunny-yj9gt I think, v = 15/4 and in decimal form it is 3.75
You are asking about this question - An object is 2cm tall is placed at 10 from a diverging mirror of focal length 6cm . Find the nature and position of the image
@@RehanaKhan-b6x if we add 1/5+1/10 it will be 3/10 which means v=10/3 also 3.33 ryt? So how is your answer is 15/4?
00:00 - Introduction
01:24 - Particle Nature Of Light
05:42 - The Wave Theory : EM Wave
07:45 - Ray Nature Of light
17:03 - Phenomenon Of Light : Reflection
19:19 - Laws Of Reflection
21:27 - Plan Mirror
21:39 - Image Formation By Plan Mirror
28:50 - Examples Of Lateral Inversion
40:25 - Important Terms : Spherical Mirrors
41:53 - Rules To Obtain Image
44:24 - Image Formation : Concave Mirror
54:52 - Image Formation : Convex Mirror
58:31 - Uses Of Mirrors
01:01:11 - Topic Sign Convention In Mirrors
01:05:33 - One Step Ahead : Formula
01:22:57 - Topic : Phenomenon Of Light : Refraction
01:26:09 - Laws Of Refraction
01:27:41 - Topic : Rules Of Refraction ( Transiting Media )
01:31:11 - When Refraction Does Not Occur !!
01:33:24 - Topic Refraction Through Glass Slab
01:38:23 - Topic Refraction Through Spherical Lenses
01:42:21 - Topic : Rules To Obtain Image
01:43:49 - Image Formation : Convex Lens
01:49:41 - Topic : Image Formation : Concave Lens
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You !
Thanks
Thanks
Just you copied from discription and pasted here 😂😊
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Description se chaap diya 😂
1:19:40
v=15/4
m=3/8
erect virtual diminished
hieght of image =0.75
Thank you sir for this useful video
1:19:44
Sir, v=30/8 cm
m=0.3 cm
Sahi ho to heart kar Dena sir🤗👍🏻👍🏻👍🏻👍🏻👍🏻
1:21:15
Sir,
isme v=3.5cm
m=0.175cm
h'=0.5cm
❤
V&E&D
BANEGI. IMG.
Height 0.75 hai
Bhaiya distance kaise nikali?
Bhai aapne galat kardiya hai, v ko simplify nahi karna tha , usse end me m ki calculation ke dauran as it is use karna tha Taaki Phir Baad me height .75 aaye aur m = .375 ho
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1:19:42 V =+ 15/4
m =+ 3/8
Hi = +0.75
Virtual and erect
diminished 👍
height of image
Mera 7.5 aa raha hai😭
Sorry pr apka ans wrong ha 15 /16 aayega =h
Mera to sahi aaya h.🎉.... Yaahoo😊
Mera bhi yahi aaya h
School ❌
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So true 😂
@@sunshine0.4class 10 board ❌ yaade ✅
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1:57:05 >
v=15cm
m= -3/2= 1.5cm
Nature of images> Enlarged,real, and inverted
Position of image> beyond 2f2
m=hi/ho
-3/2=hi/2
-6=2hi
hi=-6/2= -3cm...
Thank you!
Thank you
@@geetakharkwal8120f positive hoga kyuki convex lens h na😶
m=1.5 kaise?....mera to 1.3 aa rha hai 😢
Thank you
Bhai V 15 hai aur U -10 hai
V/u
15/-10@@mukeshtiwari6215
1:21:15
Position = 8.57cm at the back of mirror.
Nature= Erect n virtual.
Size= Diminished and lenght of image is 2.14cm.
Mara to kuch or a rha hai
@@manojkumari9220 mera bhi 😂😂😂
V =+ 60 /7 ,
Hi= 2.15....
Bhai mera v =30/8
M = 0.375
Size = .75cm
Distance kaise calculate karoon koi bta do
Abe bhai jab image diminished hai to tumhara image size ,object size we zyada kaise aara hai
1:09:55 Sir this case is of concave mirror and it is case no 4 where the object is placed between C and F and the image formed beyond C nature is real,inverted and enlarged
very good brother
❤
@@Yrgamers5 ty
✅
sahi
1:19:32
Answer→
Position of Image→v= 15/4 = 3.75 cm
Size of Image→hi= 3/4 = 0.75 cm
Nature of Image→ Virtual, Erect and Diminished
Solution in reply
As it is diverging mirror, so it is convex mirror.
Given→
Height of Object= 2 cm
u= -10 cm
f= 6 cm
Solution→
1/v + 1/u = 1/f
1/v + 1/(-10) = 1/6
1/v - 1/10 = 1/6
1/v = 1/6 + 1/10
1/v = (5+3)/30 = 8/30 = 4/15
v = 15/4 cm
m = -v/u = -(15/4)/-10
m = (15/4)/10
m = (15)/(4*10)
m = 15/40
m = 3/8 = 0.375 times
m = Hi/Ho
3/8 = Hi/2
(3/8)*2 = Hi
Hi= 3/4 = 0.75 cm
Nature of image→
As, m is positive so, the image is virtual and erect, and
As, m
Thnks but you said wrong m
@@KhushiBano-fr9oy Oh sorry
Let me edit and correct it
2:22:16
Sir isme humlog 42 na leke image aur object ka distance to diya hi hua hai to usi se nikal sakte hai na 42 lekar hi kyu nikale? Pls help me sir
@ShresthShrivastava-d3g
9 days ago
00:00 - Introduction
01:24 - Particle Nature Of Light
05:42 - The Wave Theory : EM Wave
07:45 - Ray Nature Of light
17:03 - Phenomenon Of Light : Reflection
19:19 - Laws Of Reflection
21:27 - Plan Mirror
21:39 - Image Formation By Plan Mirror
28:50 - Examples Of Lateral Inversion
40:25 - Important Terms : Spherical Mirrors
41:53 - Rules To Obtain Image
44:24 - Image Formation : Concave Mirror
54:52 - Image Formation : Convex Mirror
58:31 - Uses Of Mirrors
01:01:11 - Topic Sign Convention In Mirrors
01:05:33 - One Step Ahead : Formula
01:22:57 - Topic : Phenomenon Of Light : Refraction
01:26:09 - Laws Of Refraction
01:27:41 - Topic : Rules Of Refraction ( Transiting Media )
01:31:11 - When Refraction Does Not Occur !!
01:33:24 - Topic Refraction Through Glass Slab
01:38:23 - Topic Refraction Through Spherical Lenses
01:42:21 - Topic : Rules To Obtain Image
01:43:49 - Image Formation : Convex Lens
01:49:41 - Topic : Image Formation : Concave Lens
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You
Particle Nature Of Light
05:42 - The Wave Theory : EM Wave
07:45 - Ray Nature Of light
17:03 - Phenomenon Of Light : Reflection
19:19 - Laws Of Reflection
21:27 - Plan Mirror
21:39 - Image Formation By Plan Mirror
28:50 - Examples Of Lateral Inversion
40:25 - Important Terms : Spherical Mirrors
41:53 - Rules To Obtain Image
44:24 - Image Formation : Concave Mirror
54:52 - Image Formation : Convex Mirror
58:31 - Uses Of Mirrors
01:01:11 - Topic Sign Convention In Mirrors
01:05:33 - One Step Ahead : Formula
01:22:57 - Topic : Phenomenon Of Light : Refraction
01:26:09 - Laws Of Refraction
01:27:41 - Topic : Rules Of Refraction ( Transiting Media )
01:31:11 - When Refraction Does Not Occur !!
01:33:24 - Topic Refraction Through Glass Slab
01:38:23 - Topic Refraction Through Spherical Lenses
01:42:21 - Topic : Rules To Obtain Image
01:43:49 - Image Formation : Convex Lens
01:49:41 - Topic : Image Formation : Concave Lens
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You !
Bro u are in 10 class
Enter in class 10
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De rakha tha 😂
1:19:31 my answer is
V=30/8
M=3/8
Height of image =6/8
Nature = virtual and erect
Diminished
Its better you provide after doing all the cutting..
@@RavneetKaur-p8n yepp
Height of image 3/4
@@baburathod2766 brother meri 10th complete ho chuki h 😂
Hi 3/4
1:38:05 remember to see this while writing notes [ writing it for myself]
Thanks
Hii riya tum yaha ?😂
Numericals on concave and convex mirror
1:14:09 - object at C
1:16:46 - object at infinity
Homework question
1:19:37 - v= 15/4
m= 3/8
Erect,virtual and diminished
Object at infinity
1:21:08 - f= 15 cm
v= 60/7
m= 3/7
Erect, virtual and diminished
Object at infinity
Image ka position puccha hai to between Pole and Focus ho ga
Yr tumhara 15/4 kaise Aya mera to 30/8 h....😢
Isn't the magnification ( m) in negative = - 3/8 for the first homework question
@@janhvigupta1406you reduce it to 15/4 cuz 2(4)=8 and 2(15)= 30
@@JagmalSingh-lo1gkno it gets cut it’s positive
👑 King 👑 of physics (Rakshak sir) ❤ who agrees 👍
❤❤
And Digraj King Of SST ❤ 1:20:39
Digraj sir aap bhut hi acha padhate ho
Isi liye maine 11me humanities opt ki hai
Mere s.s.c pre board se phle 2din ka gape tha .
Usse phle mene history ki book open bhi nhi ki thi
Bss aap ke lecture dekhe the history ke or mere s.s.c me 78/80
THANK YOU SO MUCH SIR
@@Van._chaudhary.22abbe woh Asli digraj sir nahi Hai 😂
Bahrupiya Digraj sir ki nakli I'd bana ke aaya hai😂😂
Topics covered
00:00 - Introduction
01:24 - Particle Nature Of Light
05:42 - The Wave Theory : EM Wave
07:45 - Ray Nature Of light
17:03 - Phenomenon Of Light : Reflection
19:19 - Laws Of Reflection
21:27 - Plan Mirror
21:39 - Image Formation By Plan Mirror
28:50 - Examples Of Lateral Inversion
40:25 - Important Terms : Spherical Mirrors
41:53 - Rules To Obtain Image
44:24 - Image Formation : Concave Mirror
54:52 - Image Formation : Convex Mirror
58:31 - Uses Of Mirrors
01:01:11 - Topic Sign Convention In Mirrors
01:05:33 - One Step Ahead : Formula
01:22:57 - Topic : Phenomenon Of Light : Refraction
01:26:09 - Laws Of Refraction
01:27:41 - Topic : Rules Of Refraction ( Transiting Media )
01:31:11 - When Refraction Does Not Occur !!
01:33:24 - Topic Refraction Through Glass Slab
01:38:23 - Topic Refraction Through Spherical Lenses
01:42:21 - Topic : Rules To Obtain Image
01:43:49 - Image Formation : Convex Lens
01:49:41 - Topic : Image Formation : Concave Lens
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You !
Thanks bro for your help
copy and paste 😂
Hi avanit tum yaha
Introduction
01:24 - Particle Nature Of Light
05:42 - The Wave Theory : EM Wave
07:45 - Ray Nature Of light
17:03 - Phenomenon Of Light : Reflection
19:19 - Laws Of Reflection
21:27 - Plan Mirror
21:39 - Image Formation By Plan Mirror
28:50 - Examples Of Lateral Inversion
40:25 - Important Terms : Spherical Mirrors
41:53 - Rules To Obtain Image
44:24 - Image Formation : Concave Mirror
54:52 - Image Formation : Convex Mirror
58:31 - Uses Of Mirrors
01:01:11 - Topic Sign Convention In Mirrors
01:05:33 - One Step Ahead : Formula
01:22:57 - Topic : Phenomenon Of Light : Refraction
01:26:09 - Laws Of Refraction
01:27:41 - Topic : Rules Of Refraction ( Transiting Media )
01:31:11 - When Refraction Does Not Occur !!
01:33:24 - Topic Refraction Through Glass Slab
01:38:23 - Topic Refraction Through Spherical Lenses
01:42:21 - Topic : Rules To Obtain Image
01:43:49 - Image Formation : Convex Lens
01:49:41 - Topic : Image Formation : Concave Lens
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You !
1:19:45 question diverging mirror case 2
Height of image is +0.75 and magnification is +3/8
And distance of image is +15/4❤
Bhai distance of image kaise nikale ho
Bhai try kro. One time....or
@@anshjaiswal0.7 bro mera v = 30/8 aara sahi hai ?
Good bro 👍🏻
@@Boards2024Prepration bhai 30/8 divide hone k baad 15/4 hi aata h😅
00:02 Understanding the nature of light and its properties
02:08 Light is not considered a particle but rather an electromagnetic wave.
06:22 Light is an electromagnetic radiation that is visible to us.
08:28 Light travels in straight paths
12:46 Light is absorbed, transmitted, and reflected by surfaces.
15:09 Light reflection enables us to see objects
20:06 Angle of incidence and angle of reflection are always equal
22:55 Understanding reflection and refraction of light.
27:03 Understanding real and virtual images
28:55 Understanding reflection and refraction angles
34:44 Understanding focal length and radius of curvature
37:03 Understanding light reflection and refraction
41:39 Understanding rules for creating the image
45:07 Reflection and refraction of light explained through the concept of infinity in physics
51:57 Understanding the properties of convex mirrors
56:49 Understanding reflections and refractions in concave and convex mirrors
1:01:14 Understanding the sign convention of reflection and refraction on a Cartesian plane
1:03:09 Focal length of concave and convex mirrors
1:08:33 Inverted image signifies vertical formation
1:10:36 Understanding the importance of hard work and getting out of the comfort zone
1:14:56 Understanding the nature and position of M for reflection and refraction
1:18:59 Light reflection and refraction key points
1:23:56 Refraction caused by difference in optical density
1:25:48 Refraction and reflection of light
1:32:14 Understanding the concept of refraction and reflection.
1:35:11 Light undergoes lateral displacement and changes in wavelength and angle in glass
1:41:38 Understanding focal length and its effect on light
1:45:00 Understanding image formation in reflection
1:53:46 Understanding light reflection and refraction in concave and convex lenses
1:55:29 Understanding lens formula and image formation.
1:59:56 Understanding object distance and image formation in reflection and refraction.
2:01:59 Understanding power of a lens and its formula
2:06:47 Understanding refractive indices in reflection & refraction
2:08:48 Understanding refractive index and its relation to speed of light
2:12:50 Refraction of light and refractive index explained
2:15:20 Understanding absolute and relative refractive index
2:21:26 Solving for nx and discussing the answers
1:21:15 the answer is
V=60/7
Size(m)=3/7
Nature= virtual, erect,diminished
Sahi hai bhai
Bhai ye 3/7 kese aaya? Bta de...
❤same bro 😊😊
@@Devilgaming4201
Revisio ln
@@rsvchannel839- v/u ker
covered ✨
00:00 - Introduction
01:24 - Particle Nature Of Light
05:42 - The Wave Theory : EM Wave
07:45 - Ray Nature Of light
17:03 - Phenomenon Of Light : Reflection
19:19 - Laws Of Reflection
21:27 - Plan Mirror
21:39 - Image Formation By Plan Mirror
28:50 - Examples Of Lateral Inversion
40:25 - Important Terms : Spherical Mirrors
41:53 - Rules To Obtain Image
44:24 - Image Formation : Concave Mirror
54:52 - Image Formation : Convex Mirror
58:31 - Uses Of Mirrors
01:01:11 - Topic Sign Convention In Mirrors
01:05:33 - One Step Ahead : Formula
01:22:57 - Topic : Phenomenon Of Light : Refraction
01:26:09 - Laws Of Refraction
01:27:41 - Topic : Rules Of Refraction ( Transiting Media )
01:31:11 - When Refraction Does Not Occur !!
01:33:24 - Topic Refraction Through Glass Slab
01:38:23 - Topic Refraction Through Spherical Lenses
01:42:21 - Topic : Rules To Obtain Image
01:43:49 - Image Formation : Convex Lens
01:49:41 - Topic : Image Formation : Concave Lens
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You !
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1:19:42 numerical answer,
The Image is Virual, Upright and Diminished and hight of image is 0.75 or 6/8 cm2
m=3/8 i guess
@@animeheavendotcomno it's 6/ 8
@@X_euphoriax..x but i didn't deny it though
@@X_euphoriax..x height of image is 6/8 but m=3/8
M:- +3/8
Hight of image is :- +3/4
The lecture is about 2 hrs 22 mins. But it actually takes about 2 days to understand everything. To practice between the lectures... Thank you so much sir for helping us this much for free.
Really it's true
@ your pre boards over ?
29:58 The angle R2 will be 30 degrees
1:09:55 It is the case of concave mirror when the object is kept in between c and f
1:21:08 V=-10cm Nature of image is Real,Inverted and diminished
1:56:53 V=-30/7cm Nature of image is Real,Inverted and diminished
1:57:29 v=15cm,hi=3cm Natur of image is virual,erect and enlarge
1:19:36 how v is 15 can u explain
Kindly plz tell me😊
It is coming 15 by 4
@@aryavpillai4595 yeah
Brother your answer is wrong because img dist .in case of convex mirror is always +ve👍🏼
1:09:49 concave mirror case 4 b/w c and f size- enlarged nature of the image - real and inverted
FROM : PRASHANT KUMAR
1:19:31 answer is
v = 3.75cm
m = 0.375
HEIGHT OF IMAGE = 0.75
NATURE: virtual , erect , diminished .
1:21:10 ANSWER IS
V = 60/7CM
M=3/7
Height of image = 15/7
NATURE: virtual , erect , diminished.
(THANK YOU SIR FOR THE GREAT VIDEO )👍👍
How you solve thiss can u please explain! 1:19:31 question
ok
see now,
sol:-
u= -10
ho= +2
f = +6
1/f = 1/v + 1/u
so calulation ans. will 8/30=1/v
simplify it then 4/15 = 1/v
then v= 15/4
now,
m=-v/u
-15/4\-10/1
solve it
ans will m= -3/8
@@killerxgaming2491
sol
@@Ankit.Raj_2509bro you are wrong in magnification because forget to cancel ' - ' in sloving.
@@Ankit.Raj_2509Bro actually @Ayush123m is wrong because he forgot to solve the hô
M = hi/ho = -3.75 (-v)/-10 (u) then - cancelled and then 3.75 will be 0.375 the ho = 2 will be transpose to RHS and then 0.375×2 = 0.75 and this is the complete solution! Virtual+Erect+Diminished
bhai exam mai decimal mai likhna hoga kya ki fraction bhi chalega ??????
1:57:02
Size: -3cm
m: -3/2
Nature: real, inverted, enlarged
Position: 15cm
I think image is diminished because m is bigger than 1
@@Swayam_31338 if m>1 then image is enlarged. Image will be diminished if 0
Correct!
@@Swayam_31338you think wrong😂
@gamingruler4237 bhai m -3/2 he to wo kat jayega aur -1.5 hoga magnifiacton aur sara answer sahi he
Who watch this in November 😅😅😅😅
Me😅😅
I am from 9th...neev...
Aacha batt ha bhai continue😅😅@@seemakhandagale7637
Only u
@@crafty3775𝙩𝙤 𝙗𝙝𝙖𝙞 𝙩𝙪 𝙠𝙝𝙖 𝙨𝙚 𝙖𝙮𝙖
1:09:46 - Case 4 of concave mirror i.e The object will be placed b/w C and F.
1:19:43 - hi= +0.75cm ?
1:25:15 height of image +2.14 and nature virtual , errect and diminished for both questions
God job
Hi =0.85
@@premgamer0072 it's magnification's value in decimals
Yaarr mera to.... hi=0.18 a raha he.....yaarr bata do kese kiyaaa 1:19:57
Tumhari 0.75 0.85 kaise aarhi h 0.74 h kya?
✨ Topics covered ✨
00:00 - Introduction
01:24 - Particle Nature Of Light
05:42 - The Wave Theory : EM Wave
07:45 - Ray Nature Of light
17:03 - Phenomenon Of Light : Reflection
19:19 - Laws Of Reflection
21:27 - Plan Mirror
21:39 - Image Formation By Plan Mirror
28:50 - Examples Of Lateral Inversion
40:25 - Important Terms : Spherical Mirrors
41:53 - Rules To Obtain Image
44:24 - Image Formation : Concave Mirror
54:52 - Image Formation : Convex Mirror
58:31 - Uses Of Mirrors
01:01:11 - Topic Sign Convention In Mirrors
01:05:33 - One Step Ahead : Formula
01:22:57 - Topic : Phenomenon Of Light : Refraction
01:26:09 - Laws Of Refraction
01:27:41 - Topic : Rules Of Refraction ( Transiting Media )
01:31:11 - When Refraction Does Not Occur !!
01:33:24 - Topic Refraction Through Glass Slab
01:38:23 - Topic Refraction Through Spherical Lenses
01:42:21 - Topic : Rules To Obtain Image
01:43:49 - Image Formation : Convex Lens
01:49:41 - Topic : Image Formation : Concave Lens
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You !
Seedhi baat no bakwaas wali vdo h yeh !!
Thank you Rakshak sir 🤍 ~ for us you are rakshak fr 🥺🌀
Bakwas band kar bkl..
1:19:37 (v=3.75cm,m=0.375cm,hi=0.75cm)Nature of image is virtual and erect, position of image is behind the mirror, size of image is diminished (smaller than the object).
After knowing the value of v we can directly write the size of images as it is form in convex mirror and we know that convex mirror always make diminished image?
Correct I got the same answer
U divided 4/15 pagal, Ure supposed to be dividing 15 with a 4 , illiterate people.
What is hi
@@aakashveer183 height of image = 0.75cm
1:21:19 sir the height of image is 15/7 cm magnification is 3/7
(V) = 60/7
Homework completed 🎉🎉😂
Wrong answer bro
@@ayushanand6475 sahi hai
Sahi hai mere bhi itan he aaya hai
LCM aata hai ya nahi 😂😂
LCM karo 30 aaye ga tu V=30/8 hoga😂😂
27 OCTOBER 24! LIGHT DONE { js need to do some more numericals}
1:19:42
v= 15/4 cm , m = 3/8 , hi=+0.75cm. The images virtual erect and diminished.
1:22:14
v=60/7 cm , m = 3/7 , hi =15/7 cm.
1:56:47
v=-30/7 cm , 2/7=m.
1:57:02
v=15/2cm m= -3/4 and -3/2cm.
( i may be wrong somewhere or forgor to add signs ;-; )
WELZZ, TY SIR
Well done
I hope my bast friends all the students cbse board exams 95%+🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
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@@AdityaSingh-mc2oq Good Bhai
@@Pleasestudy-pm3ny Good Bhai
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Rakshak sir never found a teacher so pure and helpful like you thank you sir.....Half yearly me bhi aapki wajah se 24/26 aaye or pre board me bhi aapki wajah se marks aaye aab boards phodungi .....thank you sir love from Udaan 2.0
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Hlo ap udaan ka bhi video dekhte hai
1:19:38 ans of homework
Position of image is (v)=15 cm ans
Nature of image is (m) -1.5 hence image is real and inverted
Wrong 😅
@@TannuKumari-mk3kn how?
My also ❤❤😊
It's wrong
Object would be minus
1:57:29 v=15cm,m=-3/2,hi=-3cm
I think your wrong 😅 m=-2 / hi= -4
@@Ashok-dz2se no you are wrong
Size?
@@Ashok-dz2seyou are wrong try one more time because the answer given by @rangilal6877 is correct
@@monamamtani6063 height of the image .... negative 3 cm
Q 14 Answer :-
Given:-
Ho =+5 cm
u = -20cm
f = +15cm
Solution :
Using mirror formulae
1/f = 1/v +1/u
1/15 = 1/v - 1/20
1/20 + 1/15 = 1/v
After taking LCM
15+20/300 = 1/v
7/60 = 1/v
v = 60/7
Now,
m = -v/u
m = -60
-
7
-
-20
-
1
m = 3/7
m = hi/ho
3/7 = hi/5
15/7 = hi
hi = 2.14 cm
Therefore, height of image (hi) = 2.14cm and nature of image is virtual & errect & diminished, because (m>1)
Thankyou... ❤
Right 👍
Bhai tu iklauta shi h wrna bakio k wjh se lga ki mera hi glt h
amazinggggggg brooooooooooooo
1:19:45 size of image = 0.75cm
Image distance = 3.75cm
Position - between pole and focus,
Nature - Virtual, erect and diminished
Can u tell me how ur answer came ?
@@X_euphoriax..x given,
ho =2cm
u = -10cm
f = 6cm (convex mirror has focus on right side)
We know, 1/f = 1/v + 1/u
1/v = 1/f - 1/u
= 1/6 - (1/-10)
= 1/6 + 1/10
= 5+3/30
= 8/30
Therefore, v = 30/8 = 3.75cm
We know, m= -v/u
= -3.75/-10
= 0.375
(We get to know that the position is virtual, erect and diminished)
To find the size of image,
0.375 = hi/2 ---( m = Height of image/Height of object)
hi = 0.75cm.
Hope this helped you!
sorry the nature is virtual, erect and diminished and you will get the position by drawing the ray diagram.
@@parseries4156 thnx
Thank you bhai bahut der tak comment check kar rha tha
Aakhir me tera solution mila
Dil se dhanyavad ❤❤❤❤
1:19:17 V =30/8 , m=3/8, h =0.75
Bhai V ko cut kar sakta hai tu abhi cut kar de
H=3/8=0.375 😢
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29:56 30°
01:09:55 4) Object between centre of curvature and principal focus.
01:19:44 m = +3/8 , v = 15/4 , height of image = 3/4.
01:21:10 v = 60/7, height of image = 15/7, m = +3/7.
01:56:54 v = 15 , m = -3/2 , height of image = -3.
A very heartly thanks to rakshak sir for all this❤
m = +3/8 , v = 15/4 , height of image = 3/4 same answer
Thanks a lot ❤❤
👍
2:21:54 sir thank you for such a good and best understanding leecture
1:57:36 The size of the image is 3 cm the position of the image is 15cm in front of the mirror the image formed is real inverted and magnified
Image size will be -3cm (it will be negative )..
So sorry typing error I forgot the negative sign
Magnification is -1.5
1:09:56
Answer - Case of Concave mirror = Case number 4
Object between F & C
Beyond c hoga na bhai
Between f and c diminished hoga
Nahi bhai sahi hai
Beyond c me diminished
C pe same size
New class 10th like here
1:56:56
Yes
Ma bhi
@@girishranjan79 ya
Me
Topics covered ✨
00:00 - Introduction
01:24 - Particle Nature Of Light
05:42 - The Wave Theory : EM Wave
07:45 - Ray Nature Of light
17:03 - Phenomenon Of Light : Reflection
19:19 - Laws Of Reflection
21:27 - Plan Mirror
21:39 - Image Formation By Plan Mirror
28:50 - Examples Of Lateral Inversion
40:25 - Important Terms : Spherical Mirrors
41:53 - Rules To Obtain Image
44:24 - Image Formation : Concave Mirror
54:52 - Image Formation : Convex Mirror
58:31 - Uses Of Mirrors
01:01:11 - Topic Sign Convention In Mirrors
01:05:33 - One Step Ahead : Formula
01:22:57 - Topic : Phenomenon Of Light : Refraction
01:26:09 - Laws Of Refraction
01:27:41 - Topic : Rules Of Refraction ( Transiting Media )
01:31:11 - When Refraction Does Not Occur !!
01:33:24 - Topic Refraction Through Glass Slab
01:38:23 - Topic Refraction Through Spherical Lenses
01:42:21 - Topic : Rules To Obtain Image
01:43:49 - Image Formation : Convex Lens
01:49:41 - Topic : Image Formation : Concave Lens
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You !
Kisi ne time nhi pucha to btata kyu h
😂@@name-uq4pf
01:52:21 - Uses Of lenses
01:53:16 - Topic : One Step Ahead : Sign Convention In Lenses
01:53:55 - Topic : One Step Ahead : Formula
02:21:59 - Thank You !
1:56:53 answer will be v=-30/7, m=2/7 that means nature of the image will be diminished, virtual,erect.. position of the image will be at between f1 and O
Bro position kaise find karri?
@@RAZE.......18 U = -- 15
F = -- 6
V = - 30 / 7
M = + 2/7
VIRTUAL AND ERECT, DIMINISHED
BETWEEN F1 AND O
U is negative so it will be -2/7
1:09:53 Ans CASE 3: when object is placed between Center of curvature and focus the image will form beyong C
Characteristics :
• Real and inverted
• Beyond C
• Enlarged
• Concave Mirror
THIS LECTURE WILL HELP THE STUDENTS WHO DON'T AFFORD THE BATCHES .❤
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1:21:14 sir 🙏 f = 15 cm
V = +60/7
M _ +3/7
Hi = +15/7 cm
V and e and diminished
Also image is between O and F1
@@ImStatusZone How to find if it is between o and f1
1:56:47 v=-4.3 cm and m=0.28 Hence image will be virtual erect and diminished
1:57:29 v=15cm height of image will be 3 and nature of image will be virtual erect and enlarged
2nd answer is wrong
V=15cm
Magnification = -3/2 cm
Height of image = -3cm
Bhai 2nd answer me real aur inverted image banegi
@@BenaamBadshah-cw6un no it will be virtual and erect bcuz its +15 not -15
umm....height will be -3 and image formed will be real inverted and enlarged
@@triadictable9979 bhai 'Hi' ya m ke value pe + ya - se pata chalta hai V pe negative positive ye pata ni chlta V pe + ya - to position bata ta h image ka naa ki nature
Hw question 1:19:42 ans is the image is virtual, erect, and 3/8 times smaller than height of object.
1:18:28 = V= + 15/4 cm and Hi = + 3/4 cm , hence image is diminished, real and inverted.
Bhai Hi =3/8 cm aaraha hai
@@nikeshsingh2673 Bhai vo Hi nahi h vo m=3/8
@@thebishop_knight502are height of image se. Cross multi ply bhi to karoge to 3/4 ayega ans
Abey bhai convex mirror Mai Kab se real image form hone laga 😂
@@SleepyWerewolf-qe9jk to kisne bola real.
Virtual,erect and diminished
1:09:49 when object placed between curvature and focus in concave lens
1:31:30 wrong! , In case of normal incidence, refraction takes place and also speed of light changes but there is no deviation of light
@@Siddharth_10_ bro,you must watch the video of HC Verma regarding this topic. He is the best physics professor in india. He said that bending is only indication of refraction,but refraction is itself not bending
@@Siddharth_10_ yes bro I agree that, and that is why ncert had not touched the topic of normal incidence. Because the writers of ncert know that it would become a controversial topic.Even though in some books this topic is mentioned,but there it is written that in case of normal incidence the light goes undeviated. It isn't mentioned that refraction does not takes place.
1:19:36 ans = distance of image or v = 15/4cm. And magnification of image = 3/8. height of image = 15/2 cm and nature of image = virtual erect and diminished as its a convex mirror
Height of image galat hai
I have taken the Batch and matched from the notes he didn't missed a single topic 👍
Yes I agree😅
1:56:53
V=-4.3
M=+0.28
Virtual+ erect, diminished
As it is a diverging lens, the image formed must be virtual and erect
Answer is wrong❌❌
*virtual erect
Prr Mera toh 10 cm aya h 'V'
Mera v 15 aya
Sir thumbnail pe khus hai Orr padate wakt scerious haiii😂😂😂😂😂
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1:09:40 ..
Ans. When object is placed b/w centre of curvature (c) and Focus (f)
1:10:06 ans -ray diagram no .4 concave mirror when object is b/w C&F
Feb Preparation for board.. Attendance button ✅
Yes I agree 😅
True 😅😂
how much percentage did you get ( don't lie)
Sir in the first question 30:57 we can also asuume rays as parallel and apply co interior angles property? Because the answer will be 30 only as we know that both angles are equal.
1:57:00
V = 15 cm
m = -1.5
Hi = - 3 cm
Nature - Real & Inverted
25:27 sir head ka naam bt lo 🥺😞😭 19 Jan yaad aa jata hai😢
19 Nov bro 😢
Salute to man who compressed 8. Hr ch in two hr and 21 topo ki salami to who watch it in 2x speed😂😂😅😂
Thanks
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Haa bhai correct
Bruh Online pr 10 hrs + ka lecture ha
Kon kon 2024 2025 ke half yearly ke liye padhne aaya hai
I
I am watching in Oct 2025😊 29:17
I
@@devrajsingh6564 good
Bhai मैं state board wala hu lekin padhai ki chul ke karan mai ise dekh raha hu kyuki mene ek mahine me school ka to syllabus khatam kr diya ab cbse ki bari
1:33:13 Sir refraction occurs in the case of normal incidence but there is no bending of light. Because Refraction also means the change in speed of light.
V =15/4
m =0.37
Image formed is
Virtual and erect
Diminished
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Thankyou sir and I hope everyone get 80/80 in board as well as halfyearly
16:13 3 pero wali kursi 😢😅
1:19:43 (Position - 15/4 cm) , (Size - 3/4 cm and diminished) , (Nature - Virtual and erect)
1:19:36 My answer is -:
1. V=30/8
2. M=3/8
3. Height of object= 6/8
4. Nature -: Virtual+erect+diminished
--- 2 case of convex mirror ( between P and F )
Height of object 6/8 kaise aya?pls reply
@@AyshaAli-wr6nzhi/ho=magnification
And magnification hamne already calculate kiya tha which was equal to 3/8
Now--
hi/ho=3/8
Ho=2cm(given)
Hi/2=3/8
8hi=6
Hi=6/8=3/4
I hope u understood
1:09:55 Concave mirror in 4 th case (inlarged image, real and inverted image formed)
But waha to diverging mirror likha hai question me . Aur diverging ka matalab wo mirror convex hai ... Phir concave kaise aaya???
14:04 sir Mai toh sardiyon mein padh rha hu, Dhundh ki form mein CO2 dikh rhi hai 🤪😆😂😅😝😋
1:57:16 answer:v=15,m=-1.5,hi=-3cmreal,inverted,enlarged and beyond 2f2 4th case
Kon august 2024 me dekh. Rha h ❤😂
🤝
😶@@SamruddhiShinde-k2v
Tum 😂😂
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@@anushkaanushka4382 😁😘
no physics sir is better than rakshak sir, he is actually a saviour of every students, love you sir
Prasant sir bro
@@ranjanafaratare4188Nah bro I studied chemical equations from him I was bored instantly
1:10:13 Ans. Concave 4th case
Obj b/w C and F 😊
Rakshak sir is my favourite sir comment who agree🎉
Yes❤❤❤
Yesssssss
1:10:07 answer concave mirror ; between C and F
1:09:55 it's case no 4, where the object is in between C and F, the nature of this particular case follows as: Beyond C, Enlarged, Real, Inverted
1:9:35
homework ques
Height of object =2
distance= -10
position of image = 3.7cm
height of image = 0.75cm
nature of image = virtual and eract, diminished
hope my and is correct
Erect*
1:19:40 the nature of the image is virtual and erect, the position of the image is 3.75cm, between infinity and pole of the mirror, the size of the image is 0.75cm
Pls explain
main topics of the chapter starts here 31:10
and image formation starts here 44:30