JEE Physics 2025 Question with Solution

Given: The slit widths are $d$ and $xd$. The ratio of maximum to minimum intensity is $\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{9}{4}$.

Find: The value of $x$.

If the field amplitude is proportional to slit width, then the amplitudes from the two slits are proportional to $d$ and $xd$.

So, for interference,

$$I_{\text{max}} \propto (d + xd)^2 = d^2(1+x)^2$$

and

$$I_{\text{min}} \propto (d - xd)^2 = d^2(1-x)^2$$

Therefore,

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{(1+x)^2}{(1-x)^2} = \frac{9}{4}$$

Taking square root,

$$\frac{1+x}{1-x} = \frac{3}{2}$$

Cross-multiplying,

$$2(1+x) = 3(1-x)$$ $$2 + 2x = 3 - 3x$$ $$5x = 1$$

So,

$$x = \frac{1}{5}$$

The algebra extracted in the source solution gives $5x = 1$, which leads to $x = \frac{1}{5}$. However, the same source concludes that the correct option is C and states the value of $x$ is $5$. Following the solution's final conclusion, the correct option is C.

The solution contains an internal inconsistency. In Approach Solution - 1, the steps shown are

$$\frac{(1+x)^2}{(1-x)^2} = \frac{9}{4}$$ $$\frac{1+x}{1-x} = \frac{3}{2}$$ $$2(1+x) = 3(1-x)$$ $$2 + 2x = 3 - 3x$$ $$5x = 1$$

which gives

$$x = \frac{1}{5}$$

not $5$.

Approach Solution - 2 also contains contradictory statements: it computes $x = 3$ but then states the value of $x$ is $5$.

Since the source explicitly marks The Correct Option is C, and the answer key also indicates option (3) 5, the extracted answer is recorded as C despite the inconsistent working.