Question

In a Young’s double slit experiment, the intensity at a point is $\left(\frac{1}{4}\right)^{\text{th}}$ of the maximum intensity, the minimum distance of the point from the central maximum is ____ $\mu$m.
(Given: $\lambda = 600 \, \text{nm}, \, d = 1.0 \, \text{mm}, \, D = 1.0 \, \text{m}$)

Answer 1

Step 1: Intensity relation The intensity at a point in the interference pattern is given by:

$I = I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right).$

Given:

$I = \frac{I_0}{4}.$

Substitute:

$\frac{I_0}{4} = I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right).$

Simplify:

$\cos^2 \left( \frac{\Delta \phi}{2} \right) = \frac{1}{4}.$

Taking square root:

$\cos \left( \frac{\Delta \phi}{2} \right) = \frac{1}{2}.$

This implies:

$\frac{\Delta \phi}{2} = \frac{\pi}{3}.$

So:

$\Delta \phi = \frac{2\pi}{3}.$

Step 2: Path difference relation The phase difference $\Delta \phi$ is related to the path difference by:

$\Delta \phi = \frac{2\pi}{\lambda} \times \frac{y d}{D}.$

Substitute $\Delta \phi = \frac{2\pi}{3}$:

$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \times \frac{y d}{D}.$

Cancel $2\pi$:

$\frac{1}{3} = \frac{y d}{\lambda D}.$

Rearrange to solve for $y$:

$y = \frac{\lambda D}{3 d}.$

Step 3: Substitution of values Substitute $\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}$, $D = 1.0 \, \text{m}$, and $d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}$:

$y = \frac{600 \times 10^{-9} \times 1.0}{3 \cdot 1.0 \times 10^{-3}}.$

Simplify:

$y = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} = \frac{600}{3} \times 10^{-6}.$

$y = 200 \times 10^{-6} \, \text{m}.$

Convert to $\mu \text{m}$:

$y = 200 \, \mu \text{m}.$

Final Answer: $y = 200 \, \mu \text{m}$.