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Physics Question on Wave optics

Two slits are 1 mm apart and the screen is located 1 m away from the slits. A light wavelength 500 nm500 \, \text{nm} is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is …\dots × 10−4 m\times \, 10^{-4} \, \text{m}.

Answer

Given:
d=1 mm=10−3 m,D=1 m,λ=500 nm=5×10−7 m.d = 1 \, \text{mm} = 10^{-3} \, \text{m}, \quad D = 1 \, \text{m}, \quad \lambda = 500 \, \text{nm} = 5 \times 10^{-7} \, \text{m}.
For the central maximum of the single-slit pattern to contain 10 maxima of the double-slit pattern:
10×λDd=2λDa,10 \times \frac{\lambda D}{d} = \frac{2\lambda D}{a},
where aa is the slit width.
Rearranging for aa:
a=d5.a = \frac{d}{5}.
Substitute d=10−3 md = 10^{-3} \, \text{m}:
a=10−35=2×10−4 m.a = \frac{10^{-3}}{5} = 2 \times 10^{-4} \, \text{m}.
Thus, the slit width is:
a=2×10−4 m.a = 2 \times 10^{-4} \, \text{m}.

Explanation

Solution

Given:
d=1 mm=10−3 m,D=1 m,λ=500 nm=5×10−7 m.d = 1 \, \text{mm} = 10^{-3} \, \text{m}, \quad D = 1 \, \text{m}, \quad \lambda = 500 \, \text{nm} = 5 \times 10^{-7} \, \text{m}.
For the central maximum of the single-slit pattern to contain 10 maxima of the double-slit pattern:
10×λDd=2λDa,10 \times \frac{\lambda D}{d} = \frac{2\lambda D}{a},
where aa is the slit width.
Rearranging for aa:
a=d5.a = \frac{d}{5}.
Substitute d=10−3 md = 10^{-3} \, \text{m}:
a=10−35=2×10−4 m.a = \frac{10^{-3}}{5} = 2 \times 10^{-4} \, \text{m}.
Thus, the slit width is:
a=2×10−4 m.a = 2 \times 10^{-4} \, \text{m}.