In a Young's double-slit experiment, the slit separation is 1.5 mm. Twenty interference maxima are contained within the central maximum of the single-slit diffraction envelope. If the width of each slit is x×10⁻³ cm, find x.
Physics · Wave Optics · young's dubble slit experiments
JEE Mains 2025 4 April Evening Shift

Question

In a Young's double-slit experiment, the slit separation is 1.5 mm. Twenty interference maxima are contained within the central maximum of the single-slit diffraction envelope. If the width of each slit is x×10⁻³ cm, find x.

Physics · Wave Optics · young's dubble slit experiments

JEE Mains 2025 4 April Evening Shift

G1 Physics · Accepted Answer

The width of the central single-slit diffraction maximum on the screen is W = 2λD/a, where a is the slit width. The fringe width of the double-slit interference pattern is β = λD/d, where d is the slit separation. If 20 maxima are contained within the central diffraction maximum, the source solution uses W = 20β. Therefore, 2λD/a = 20(λD/d). Cancel λD: 2/a = 20/d ⇒ a = d/10. Given d = 1.5 mm = 1.5×10⁻³ m, a = 1.5×10⁻⁴ m = 0.015 cm = 15×10⁻³ cm. Thus x = 15.