Q75 / 90 · JEE Main 2019 (10 Apr Shift 2) · Physics

In a Young's double-slit experiment, the ratio of the slit's width is $4 : 1$ . The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

A
$25 : 9$
Correct answer
B
$9 : 1$
C
${(\sqrt{3} + 1)}^{4} : 16$
D
$4 : 1$

Solution

I n t e n s i t y

\propto

w i d t h o f s l i t

\frac{I_{1}}{I_{2}} = \frac{4}{1}

.

\frac{I_{m a x}}{I_{m i n}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\sqrt{\frac{I_{1}}{I_{2}}} - 1})}^{2}

= {(\frac{\sqrt{4} + 1}{\sqrt{4} - 1})}^{2} = {(\frac{3}{1})}^{2} = \frac{9}{1}

I_{m a x} : I_{m i n} = 9 : 1

Hence, the correct answer is option: 'B'.

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