Question

In Young’s double slit experiment, light from two identical sources is superimposing on a screen. The path difference between the two lights reaching a point on the screen is $\frac{7\lambda}{4}$. The ratio of the intensity of the fringe at this point with respect to the maximum intensity of the fringe is:

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{1}{2}$

Answer 2

Given:
- Path difference $\Delta x = \frac{7\lambda}{4}$.

Step 1. Calculate the phase difference $\phi$:

$\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \times \frac{7\lambda}{4} = \frac{7\pi}{2}$

Step 2. Determine the intensity at this point:
The intensity $I$ at a point with phase difference $\phi$ is given by:

$I = I_{\text{max}} \cos^2 \left(\frac{\phi}{2}\right)$
Step 3. Calculate $\frac{I}{I_{\text{max}}}$:

$\frac{I}{I_{\text{max}}} = \cos^2 \left(\frac{\phi}{2}\right) = \cos^2 \left(\frac{7\pi}{4}\right)$

Using $\cos \frac{7\pi}{4} = \cos \left(2\pi - \frac{\pi}{4}\right) = \cos \frac{\pi}{4}$, we get:

$\frac{I}{I_{\text{max}}} = \cos^2 \frac{\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$

Thus, the ratio of intensity at this point to the maximum intensity is .$\frac{1}{2}$

The Correct Answer is: $\frac{1}{2}$