A glass plate is placed above a glass cube of 2cm edges in such a way that there remains a thin air layer of thickness ‘d’ between them (see figure). Electromagnetic radiation of wavelength between 400nm and 1150nm (for which the plate is penetrable) is incident perpendicular to the plate from above, is reflected from both air surfaces (bottom of the glass plate and top of the cube) and interferes. In this range, only two wavelengths give maximum reinforcements, one of them is 400nm. $λ$ is the second wavelength. T is change in temp. necessary to warm up the cube so as it would touch the plate. The coefficient of linear thermal expansion of material of cube is $α = 8 \times 10^{- 6} / ° C,$ the refractive index of the air $μ = 1.$ The distance of the bottom of the cube from the plate do not change during warming up. Then

Question

A glass plate is placed above a glass cube of 2cm edges in such a way that there remains a thin air layer of thickness ‘d’ between them (see figure). Electromagnetic radiation of wavelength between 400nm and 1150nm (for which the plate is penetrable) is incident perpendicular to the plate from above, is reflected from both air surfaces (bottom of the glass plate and top of the cube) and interferes. In this range, only two wavelengths give maximum reinforcements, one of them is 400nm.

λ

is the second wavelength. T is change in temp. necessary to warm up the cube so as it would touch the plate. The coefficient of linear thermal expansion of material of cube is

α = 8 \times 10^{- 6} / ° C,

the refractive index of the air

μ = 1.

The distance of the bottom of the cube from the plate do not change during warming up. Then
Question Image

Accepted Answer

2μd=(2n+1)λ/2 for constructive interference λ=4μd2n+1λ=4μd,4μd3,4μd5,4μd7......As n↑λ↓ 400=4μa2n+1⇒4μd=400(2n+1) ....(1) Again λ=4μd2(n−1)+1 for next one λ<1150⇒4μd(2n+1)−2<1150⇒400(2n+1)(2n+1)−2<1150⇒(2n+1)(1150−400)>2300 2n+1>23075 n>3130 ∴n=2,3,4...... λ>1150 for (n−2)⇒4μd2(n−2)+1>1150⇒n<7730⇒n=0,1,2 n=2(common)⇒4×1×d5=400⇒d=500 nm ΔL=laT⇒T=dha=3.125°C and λ=4μd2(2−1)+1=4×1×5003=666.67nm

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App