1. A double slit diffraction pattern has an intensity envelope around the centra

Question

1. A double slit diffraction pattern has an intensity envelope around the central maximum. And inside it there are n fringes. If the wavelength is 420nm and d=0.200 nm and a = 45 micrometers, then what is the ratio of the intensity of the central fringe compared to the third fringe?(This should be bigger than one.)

2. Young’s Experiment is performed with wavelength 420nm and the slits are 2mm apart. How far away should you place a screen so that the closest fringes are 2mm apart?

3. You wish to coat some glass(n=1.6) with a thin film of transparent material (n=1.33)so that light of wavelength 420nm (in vacuum) does not reflect. What minimum thickness (in nm) do you need?

Explanation / Answer

in double slit diffraction

I = Im*(cos V)^2*((sin A)/A)^2

A = pi*a*(sin theta)/lambda

B = pi*d*(sin theta)/lambda

Im = intensity of the center pattern

for the third fringe

m*lambda = d*sin theta

sin theta = 3*lambda/d

using above equation

B = 3*pi

A = 3*pi*a/d = 3*pi*45*10^-6/(0.2*10^-3) = 0.675*pi

(cos B)^2 = 1

(sin A)/A)^2 = [((sin (0.675*pi deg))/(0.675)]^2 = 0.1616

I/Im = 0.1616

Now you need

Im/I = 1/0.1616 =6.18

Let me know if it's still wrong.

I think that this answer is wrong, because value of d is very very low, please recheck the value of d and let me know if it's wrong.

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