Question

In a YDSE, width of one slit is three times the other. Amplitude is proportional to slit-width. Find the ratio $I_{max}/I_{min}$.

• A. 4 : 1
• B. 2 : 1
• C. 1 : 4
• D. 3 : 1

Hint:

Intensity $I \propto A^2$ and $A \propto \text{width}$. Therefore $I \propto \text{width}^2$. Be careful, as some questions state $I \propto \text{width}$, but here $A \propto \text{width}$ is specified.

Answer 1

The Correct Option is A

Step 1: $w_1 = 3w_2$. Given $A \propto w$, so $A_1 = 3A_2$.
Step 2: $I_{max} = (A_1 + A_2)^2 = (3A_2 + A_2)^2 = (4A_2)^2 = 16A_2^2$.
Step 3: $I_{min} = (A_1 - A_2)^2 = (3A_2 - A_2)^2 = (2A_2)^2 = 4A_2^2$.
Step 4: Ratio $I_{max}/I_{min} = 16/4 = 4 : 1$.