QJEEM25Ph730a: JEE Main 2025 Physics - Class 12 JEEM25Ph730a Solution: The width of one of the two slits in Young's double slit…

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

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

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 $x$ :

$2 + 2x + 3 = 3x \quad \Longrightarrow \quad 5 + 2x = 3x,$

which implies

$x = 5.$

Thus, the value of $x$ is

$\boxed{5}.$