JEE Main 2021 — Wave Optics Question with Solution

Given, b₁ = 3b₂

Here, b₁ = width of the one of the two slit and b₂ = width of the other slit.

As we know that,

Intensity, I $$\propto$$ (Amplitude)²

$$\Rightarrow$$ $$\frac{I_1}{I_2}={\left(\frac{b_1}{b_2}\right)}^2\Rightarrow \frac{I_1}{I_2}={\left(\frac{3b_2}{b_2}\right)}^2$$

$$I_1=9I_2$$

As we know, the ratio of the minimum intensity to the maximum intensity in the interference pattern,

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

Substituting the values in the above equations, we get

$$\frac{I_{\min }}{I_{\max }}=\frac{{(\sqrt{9I_2}-\sqrt{I_2})}^2}{{(\sqrt{9I_2}+\sqrt{I_2})}^2}={\left(\frac{3-1}{3+1}\right)}^2$$

$$\frac{I_{\min }}{I_{\max }}=\frac{1}{4}$$

Comparing with, $$\frac{I_{\min }}{I_{\max }}=\frac{x}{4}$$

The value of x = 1.