dear mam, please tell me that ,in the last question why are you taking the path differnece 2utcosr ? , for reflected light the effective path differnce for brightness is 2utcosr +lamda/2 you took the path differnce for transmitted light (brightness)that is 2utcosr =(m)lamda please clear my doubt mam please reply
mam in the 2nd question is compulsory to convert angstrong to meter ? also didnt understand why in the 3rd question the wave incident normally so r must be 90 deg
In thi video I got only numericals ...No waste of time ... Really appreciated .. thank u mam 🙏🙏🙏😊
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"Thank you so much for your patience and guidance in helping me understand my weakest subject and I'm truly grateful
Love from Maharashtra ❤
Thank you Gayatri. All the best
Mam aap bhut accha padate ho thankyou mam
Thank you too
mam your teaching style is very good please upload all the chapter of physics its my humple request
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Thanks madam sir says only important problem
But you tell every problem in thin film this problems so helps to students😀🙂
At 5:48 u consider it as bright. Why mam ?
Because they mention visible region so that's why u consider it or any other reason
32:10 mam for brightness condition is 2 mu t cos(r) = (2n-1)lamda/2
But why u take dark condition pls clarify mam
dear mam,
please tell me that ,in the last question why are you taking the path differnece 2utcosr ? , for reflected light the effective path differnce for brightness is 2utcosr +lamda/2
you took the path differnce for transmitted light (brightness)that is 2utcosr =(m)lamda
please clear my doubt mam
please reply
Mam please solve questions on Fresnel biprism
This is very useful .thank you madam😍😍
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A lot of thanks ma'am . It is really helpful
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Thankyou so much mam and keep it up for this wonderful content✌️
Do u have idea about cosr =1 pliz
@@Sheikh_Nishad_Hyder1948 i=r, when their is nothing in question about i value we consider i=r=0, cos(0)=1, proved cosr =1.
wonderful lecture , plenty of numericals ,great explanations maam
going to give tutorial tomorrow on 26/1/23 in dj on this chapter
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Wonderful explanation
thnq for a great explaination mam...
Great video
Thank you
My doubts got cleared mam....
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Thank you mam it's very helpful
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thank you so so much mam I am searching for this video and i highly required it🙏🙏
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Great lecture mam❤️🤝🏻
Very effective session ❤❤
Glad to hear that
What is the wavelength range of visible region?
It's 4000-7000 A°
@@sohailshaikh5224 ok thank you
4000-7000
Thanks for helping me
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Thanku ma'am
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Lots of thanks mam ❤
Thank you too. Best wishes
Thank YOU! So much ma'am
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Thank you.
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mam why did u not convert the wavelength in the first question for t minimum
Very helping 🤩
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Veey helpful madam. 🖖🖖
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Thank you so much ma'am 🙏
Why in brightness cosr =1 where it's come from???
Because r= 0 since i= 0 when light falls normally
i want u in college in which i m studying so badly 😊
Thank you .
mam in the 2nd question is compulsory to convert angstrong to meter ?
also didnt understand why in the 3rd question the wave incident normally so r must be 90 deg
Not compulsory. You can write answer in Armstrong too
For normal incidence of light,, the refracted ray has no deviation. So angle of refraction is zero degree.
@@physicsjessy7746 got it mam .
video is not clear mam
Thank you mam
Keep watching