Question

In Young's double slit experiment the two coherent sources have different amplitudes. If the ratio of maximum intensity to minimum intensity is 16 : 1, then the ratio of amplitudes of the two sources will be ______.

• A. 4 : 1
• B. 5 : 3
• C. 1 : 4
• D. 1 : 16

Hint:

For interference, use \( \frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 \) to find amplitude ratios.

Answer 1

The Correct Option is B

In Young's double slit experiment, for amplitudes \( a_1 \) and \( a_2 \), intensity \( I \propto a^2 \).
\[ I_{\text{max}} = (a_1 + a_2)^2, \quad I_{\text{min}} = (a_1 - a_2)^2. \] \[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = 16 \quad \Rightarrow \quad \frac{a_1 + a_2}{a_1 - a_2} = 4. \] Let \( r = \frac{a_1}{a_2} \). Then: \[ \frac{a_1 + a_2}{a_1 - a_2} = \frac{r + 1}{r - 1} = 4. \] \[ r + 1 = 4(r - 1) \quad \Rightarrow \quad r + 1 = 4r - 4 \quad \Rightarrow \quad 5 = 3r \quad \Rightarrow \quad r = \frac{5}{3}. \] Answer: 5 : 3.