Question

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. $9:1$
• B. $(\sqrt{3}+1{)}^4:16$
• C. $25:9$
• D. $4:1$

Answer 1

The correct option is A $9:1$

Given:
Ratio of the slit's width $=4:1$

Let $I_1,I_2$ are intensity of light coming from two slits.

$As,I\propto W$

$\therefore \frac{I_1}{I_2}=\frac{4}{1}$

$I_1=4I_0$

$I_2=I_0$

Now, ratio of maximum to minimum Intensity is given as
$\frac{I_{max}}{I_{min}}=\frac{(\sqrt{I_1}+\sqrt{I_2}{)}^2}{(\sqrt{I_1}-\sqrt{I_2}{)}^2}={\left(\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right)}^2={\left(\frac{\sqrt{4}+1}{\sqrt{4}-1}\right)}^2=\frac{9}{1}$

Hence option $\left(B\right)$ is correct.