The width of one of the two slits in a Young's double slit...

The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is:

Options

A $1: 1$
B $4: 1$
C $9: 1$
D $16: 1$

Correct answer

C. $9: 1$

Step-by-step solution

Since, Intensity

\propto

width of slit

(\omega)

\begin{aligned} & \text{ so, } I_1=I, I_2=4 \mathrm{I} \\ & I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=I \\ & I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=9 \mathrm{I} \\ & \frac{I_{\max }}{I_{\min }}=\frac{9 \mathrm{I}}{\mathrm{I}}=\frac{9}{1} \end{aligned}

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