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 central maximum of the single slit diffraction pattern, then the width of each slit is x × 10⁻³ cm, where x-value is _________ .

Solution

Physics sir · 4 steps

Fit 20 fringe widths exactly into the full central diffraction maximum; λ and D cancel out.

1. Two widths. The double-slit fringe width is β = (λ D)/d. The central maximum of single-slit diffraction runs from the first minimum on one side to the first minimum on the other, so its full width is (2λ D)/a.
2. Set them equal. 20 maxima inside the central maximum means 20 fringe widths fill its full width.

   20 × (λ D)/d = (2λ D)/a ⇒ a = 2d/20 = d/10

3. Put in numbers. λ and D cancel, so only d matters. a = (1.5 mm)/10 = 0.15 mm = 0.015 cm = 15 × 10⁻³ cm.
4. So. The width of each slit is 15 × 10⁻³ cm, so x = 15.

The common slip: 7.5. Taking the central maximum width as (λ D)/a (only one side of the centre) gives a = d/20 = 7.5 × 10⁻³ cm, not 15.

The idea to keep. For n interference maxima inside the central diffraction maximum, use nλ/d = 2λ/a, so a = 2d/n. Always take the full width of the central maximum, on both sides of the centre.

Answer from the NTA final answer key. Solution by PrepWiser.