JEE Main 2025 Jan 23 Shift 2, Physics Q13: Young’s Double Slit Experiment

JEE Main2025Jan 23, Shift 2Physics

The width of one of the two slits in Young's double slit experiment is $d$ while that of the other slit is $x d$ . If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is $9 : 4$ then what is the value of $x$ ?

(Assume that the field strength varies according to the slit width.)

A
$2$
B
$3$
C
$5$
D
$4$

Solution

If the field amplitude is proportional to slit width, then $a_{1} \propto d$ and $a_{2} \propto x d$ . Intensities scale as the square of amplitudes:

$I_{1} \propto d, I_{2} \propto x d .$

The interference extremes are:

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

Take the positive square root: $\frac{I _{1} + I _{2}}{∣ I _{1} - I _{2} ∣} = \frac{3}{2}$ .

Substitute the amplitude ratio (assume $x > 1$ ): $\frac{( x + 1 ) d}{( x - 1 ) d} = \frac{3}{2}$ .

$2 (x + 1) = 3 (x - 1) \Rightarrow 2 x + 2 = 3 x - 3 \Rightarrow x = 5.$

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JEE Main 2025 Jan 23 Shift 2, Physics Q13: Young’s Double Slit Experiment

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