Question

Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is (a) 500 nm (visible light); (b) 50 μm (infrared radiation); (c) 0.500 nm (X-rays)?

Answer 1

Given : 2W = 6mm W = 3mm = 3 x 10⁻³ m, y=2.5 m,

(a) λ₁ = 500 nm = 5 x 10⁻⁷ m

(b) λ₂ = 50 μm = 5 x 10⁻⁵ m

(c) λ₃ = 0.500 nm = 5 x 10⁻¹⁰ m

Let a be the slit width.

(a) W = \(\frac{yλ_1}{a}\)

∴ a = \(\frac{yλ_1}{W}\) = \(\frac{(2.5)(5×10^{-7})}{3×10^{-3}}\)

= 4.167 x 10⁻⁴ m

= 0.4167 mm

(b) W =\(\frac{yλ_2}{a}\)

∴ a = \(\frac{yλ_2}{W}\)=\(\frac{(2.5)(5×10^{-10})}{3×10^{-3}}\)

= 4.167 x 10⁻² m

= 41.67 mm

(C) W = \(\frac{yλ_3}{a}\)

∴ a = \(\frac{yλ_3}{W}\)=\(\frac{(2.5)(5×10^{-5})}{3×10^{-3}}\)

= 4.167 x 10⁻⁷ m

= 4.167 x 10⁻⁴ mm