JEE Main 2025 — Wave Optics Question with Solution

Let the amplitude from the slit of width $$d$$ be proportional to $$E$$ and from the slit of width $$xd$$ be proportional to $$xE$$ (with $$x>1$$, as is evident from the given options).

For two coherent waves with amplitudes $$E_1$$ and $$E_2$$, the resultant intensity when added in phase (maximum) and out of phase (minimum) is given by

$$I_{\max }\propto (E_1+E_2{)}^2\text{and}{I}_{\min }\propto (E_2-E_1{)}^2.$$

Here, setting

$$E_1=E$$ (from the narrower slit of width $$d$$) and

$$E_2=xE$$ (from the wider slit of width $$xd$$),

we have

$$I_{\max }\propto (E+xE{)}^2=E^2(1+x{)}^2,$$

and since $$x>1$$ the subtraction in the destructive case gives

$$I_{\min }\propto (xE-E{)}^2=E^2(x-1{)}^2.$$

The ratio of maximum to minimum intensity is therefore

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

We are given that

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

Taking the square root of both sides (noting that all quantities are positive) leads to

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

To solve for $$x$$, cross-multiply:

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

Expanding both sides:

$$2+2x=3x-3.$$

Rearrange the equation to isolate $$x$$:

$$2+2x+3=3x⟹5+2x=3x,$$

which implies

$$x=5.$$

Thus, the value of $$x$$ is

$$\boxed{5}.$$