JEE Main 2025 Apr 4 Shift 2, Physics Q24: Young’s Double Slit Experiment

JEE Main2025Apr 4, Shift 2Physics

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is $x \times 1 0^{- 3}$ cm, where $x$ -value is ______.

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Solution

The fringe width of the double-slit pattern is $β = \frac{λ D}{d}$ . So the linear span of 20 fringes is $\frac{20 λ D}{d}$ .

The half-width of the central maximum of the single-slit pattern (slit width $a$ ) is $\frac{λ D}{a}$ , so the full width is $\frac{2 λ D}{a}$ .

Setting these equal:

$\frac{20 λ D}{d} = \frac{2 λ D}{a} \Rightarrow a = \frac{d}{10}$ .

With $d = 1.5 mm = 1.5 \times 1 0^{- 1} cm$ :

$a = \frac{1.5 \times 1 0 ^{- 1}}{10} = 1.5 \times 1 0^{- 2} cm = 15 \times 1 0^{- 3} cm$ .

$∴ x = 15$ .

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JEE Main 2025 Apr 4 Shift 2, Physics Q24: Young’s Double Slit Experiment

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