Question

Two slits in Young's interference experiment have width in the ratio 1: 4. The ratio of intensity at the maxima and minima in their interference is ________.

• A. $\frac{1}{9}$
• B. $\frac{1}{4}$
• C. $\frac{9}{1}$
• D. $\frac{2}{4}$

Hint:

Slit width $\propto$ Intensity $\propto$ (Amplitude)$^{2}$.

Answer 1

The Correct Option is C

Step 1: Concept
Slit width ratio $w_{1}/w_{2}$ is equal to the intensity ratio $I_{1}/I_{2}$, and intensity $I \propto A^{2}$.
Step 2: Analysis
$I_{1}/I_{2} = 1/4 \Rightarrow (A_{1}/A_{2})^{2} = 1/4 \Rightarrow A_{2} = 2A_{1}$.
Step 3: Calculation
$\frac{I_{max}}{I_{min}} = \frac{(A_{1} + A_{2})^{2}}{(A_{1} - A_{2})^{2}} = \frac{(A_{1} + 2A_{1})^{2}}{(A_{1} - 2A_{1})^{2}} = \frac{(3A_{1})^{2}}{(-A_{1})^{2}} = \frac{9}{1}$.
Step 4: Conclusion
Hence, the ratio is 9:1.
Final Answer:(C)