In Young's double slit experiment, one of the slit is wider than...

In Young's double slit experiment, one of the slit is wider than other, so that the 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

is given by

Options

A $\frac{I_m}{9}(4+5 \cos \phi)$
B $\frac{\mathrm{l}_{\mathrm{m}}}{3}\left(1+2 \cos ^2 \frac{\phi}{2}\right)$
C $\frac{I_m}{5}\left(1+4 \cos ^2 \frac{\phi}{2}\right)$
D $\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right)$

Correct answer

D. $\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right)$

Step-by-step solution

Let

A_1=A_0, A_2=2 A_0

If amplitude of resultant wave is

A

then

\mathrm{A}^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2 \cos \phi

For maximum intensity,

\mathrm{A}_{\max }^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2

\therefore\frac{A^2}{A_{\max }^2}=\frac{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}{A_1^2+A_2^2+2 A_1 A_2}

=\frac{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right) \cos \phi}{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right)}

\frac{

\mathrm{I}

}{

\mathrm{I}

_{\mathrm{m}}}=\frac{5+4 \cos \phi}{9}=\frac{1+8 \cos ^2(\phi / 2)}{9}

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