Question

Slit width in a Young's double slit experiment are in the ratio $9:4$. The ratio of intensity at minima to that at maxima, is

• A. $4:9$
• B. $16:81$
• C. $1:25$
• D. $1:5$

Answer 1

The correct option is C $1:25$

As width of a slit is directly proportional to its intensity,
$\Rightarrow \frac{I_1}{I_2}=\frac{9}{4}$

Also $\frac{I_{\text{min}}}{I_{\text{max}}}=\frac{(\sqrt{I_1}-\sqrt{I_2}{)}^2}{(\sqrt{I_1}+\sqrt{I_2}{)}^2}...\left(1\right)$

$\Rightarrow \frac{I_{\text{min}}}{I_{\text{max}}}=\frac{{[\sqrt{\left(\frac{I_1}{I_2}\right)}-1]}^2}{{[\sqrt{\left(\frac{I_1}{I_2}\right)}+1]}^2}$

$=\frac{{(\frac{3}{2}-1)}^2}{{(\frac{3}{2}+1)}^2}$

$\therefore \frac{I_{\text{min}}}{I_{\text{max}}}=\frac{1}{25}$

Hence, $\left(C\right)$ is the correct answer.

Why this question? To understand the relation between width of a slit and maximum and minimum intensity of light.