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Wave Optics Part 1

Section: Physics

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Question : 89 of 100

Marks: +1, -0

In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If

I_m

be the maximum intensity, the resultant intensity I when they interfere at phase difference

\phi

$\;\frac{I_m}{9}(4+5 \cos \phi)$
$\;\frac{I_m}{3}\left(1+2 \cos^2 \frac{\phi}{2}\right)$
$\;\frac{I_m}{3}\left(1+4 \cos^2 \frac{\phi}{2}\right)$
$\;\frac{I_m}{9}\left(1+8 \cos^2 \frac{\phi}{2}\right)$

Solution:

Let

a_1=a, I_1=a_1^2=a^2

a_2=2a, I_2=a_2^2=4a^2

Therefore

I_2=4I_1

\;I_r=I_1+I_2+2\sqrt{I_1 I_2 \cos \phi}

\;I_r=I_1+4I_1+2\sqrt{4 I_1^2} \cos \phi

\;\Rightarrow I_r=5I_1+4I_1 \cos \phi

......(1)

Now,

I_{\max} = (a_1+a_2)^2 = (a+2a)^2 = 9a^2

I_{\max}=9I_1 \Rightarrow I_1 = \frac{I_{\max}}{9}

Substituting in equation (1)

I_r=\;\frac{5 I_{\max}}{9}+\;\frac{4 I_{\max}}{9}\cos \phi

I_r=\;\frac{I_{\max}}{9}[5+4 \cos \phi]

I_r=\;\frac{I_{\max}}{9}\left[5+8 \cos^2 \frac{\phi}{2}-4\right]

I_r=\;\frac{I_{\max}}{9}\left[1+8 \cos^2 \frac{\phi}{2}\right]

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