Question

The width of one of the two slits in Young's double-slit experiment is \( d \) while that of the other slit is \( x d \). If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9 : 4, then what is the value of \( x \)? (Assume that the field strength varies according to the slit width.)

• A. 2
• B. 3
• C. 5
• D. 4

Hint:

In interference patterns, the intensity ratio depends on the square of the ratio of slit widths. Use this relation to solve for unknowns in similar problems.

Answer 1

The Correct Option is C

In Young’s double-slit experiment, the intensity at the maxima and minima is related to the slit widths. The maximum intensity is proportional to \( I_{\text{max}} \sim I_0 \) and the minimum intensity is proportional to \( I_{\text{min}} \sim I_0 \cdot \left(\frac{d}{x d}\right)^2 \). The ratio of the maximum to minimum intensity is given by: \[ \frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{x}{1} \right)^2 = 9 \quad \Rightarrow \quad x = 3. \] Thus, the value of \( x \) is 5.