Question

In Young’s double slit experiment the ratio of intensity of the maxima and minima in the interference experiment is 25:9 The ratio of widths of two slits is:

• A. 18:3
• B. 4:1
• C. 8:1
• D. 16:1

Answer 1

Answer: (D)

16:1

Answer 2

: Here , $\frac{I_{\max}}{I_{\min} = \frac{25}{9}}$

Or $\left( \frac{A_{1} + A_{2}}{A_{1} - A_{2}} \right)^{2} = \frac{25}{9}$

$\frac{\frac{A_{1}}{A_{2}} + 1}{\frac{A_{1}}{A_{2}} - 1} = \frac{5}{3} \Rightarrow \frac{A_{1}}{A_{2}} = 4$

$\therefore$width ratio of two slits

$\frac{d_{1}}{d_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}} = \frac{16}{1} = 16:1$