Solved problems on interference in thin film

[@aditirocks9215]

In thi video I got only numericals ...No waste of time ... Really appreciated .. thank u mam 🙏🙏🙏😊

[@physicsjessy7746]

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[@GayatriBhanage]

"Thank you so much for your patience and guidance in helping me understand my weakest subject and I'm truly grateful Love from Maharashtra ❤

[@physicsjessy7746]

Thank you Gayatri. All the best

[@5Finder]

mam your teaching style is very good please upload all the chapter of physics its my humple request

[@physicsjessy7746]

🙏🙏

[@vivekdesu5007]

Thanks madam sir says only important problem But you tell every problem in thin film this problems so helps to students😀🙂

[@UdayRajofficialvlog91]

Mam aap bhut accha padate ho thankyou mam

[@physicsjessy7746]

Thank you too

[@AjitKumar-b6b5g]

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

[@Faithonurself]

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

[@sasindumihiram3944]

This is very useful .thank you madam😍😍

[@physicsjessy7746]

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[@SachinKumar-le2qj]

Great video

[@physicsjessy7746]

Thank you

[@Faithonurself]

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

[@since-now]

bcz its given that strongly reflect

[@bubaibej6858]

A lot of thanks ma'am . It is really helpful

[@physicsjessy7746]

🙏🙏

[@rick2342]

Thankyou so much mam and keep it up for this wonderful content✌️

[@Sheikh_Nishad_Hyder1948]

Do u have idea about cosr =1 pliz

[@rick2342]

@@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.

[@abhaysrivastava-musical1300]

Mam please solve questions on Fresnel biprism

[@shreyatech00100]

Great lecture mam❤️🤝🏻

[@rohitreddy1340]

My doubts got cleared mam....

[@physicsjessy7746]

🙏

[@suryanshsingh5304]

thank you so so much mam I am searching for this video and i highly required it🙏🙏

[@physicsjessy7746]

🙏

[@mustakimmirza9036]

Lots of thanks mam ❤

[@physicsjessy7746]

Thank you too. Best wishes

[@chaitanyalimaye4619]

What is the wavelength range of visible region?

[@sohailshaikh5224]

It's 4000-7000 A°

[@chaitanyalimaye4619]

@@sohailshaikh5224 ok thank you

[@yashdalavi401]

4000-7000

[@rohan8767]

mam why did u not convert the wavelength in the first question for t minimum

[@anirbandey6206]

Veey helpful madam. 🖖🖖

[@physicsjessy7746]

🙏

[@husaincementwala]

wonderful lecture , plenty of numericals ,great explanations maam going to give tutorial tomorrow on 26/1/23 in dj on this chapter

[@physicsjessy7746]

🙏

[@udayunnikrishnan6702]

Very effective session ❤❤

[@physicsjessy7746]

Glad to hear that

[@kanaiyakanabar5178]

Wonderful explanation

[@manaschoudhari380]

Thank you mam it's very helpful

[@physicsjessy7746]

🙏🙏

[@shivpujansingh1053]

Very helping 🤩

[@physicsjessy7746]

🙏

[@OnkarAvhad-nm7dl]

In question 1 for minimum thickness you took m=1 but for that formula 2utcosr=(2m+1)lamda/2 the value of m should start from 0 and the minimum value for thickness is m=0 And for the formula 2utcosr=(2m-1)lamda/2 the minimum value m should start from 1... Why you took 1 the answer is wrong Am I right mam Pls clear my doubt

[@Sheikh_Nishad_Hyder1948]

Why in brightness cosr =1 where it's come from???

[@indra7018]

Because r= 0 since i= 0 when light falls normally

[@Nithya-cj6pb]

thnq for a great explaination mam...

[@aftabahmed9636]

Thank YOU! So much ma'am

[@physicsjessy7746]

🙏

[@kennnyyyyyy11]

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

[@physicsjessy7746]

Not compulsory. You can write answer in Armstrong too

[@physicsjessy7746]

For normal incidence of light,, the refracted ray has no deviation. So angle of refraction is zero degree.

[@kennnyyyyyy11]

@@physicsjessy7746 got it mam .

[@akashkekan9707]

Thanks for helping me

[@physicsjessy7746]

🙏

[@ManavMSanger]

Thank you.

[@physicsjessy7746]

🙏

[@Shahzadahmed2016vs]

Thank you so much ma'am 🙏

[@yashashdahbara1869]

Thankyou

[@petrosukums2510]

❤

[@dshikha6516]

Thanku ma'am

[@physicsjessy7746]

🙏🙏

[@RohitKumarSingh-cu8ed]

i want u in college in which i m studying so badly 😊

[@physicsjessy7746]

Thank you .

[@AKEPOGUBLESSY24BCC706]

video is not clear mam

[@ummesalma6856]

Thank you mam

[@physicsjessy7746]

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