A single slit of width 0.1 mm is illuminated by a mixture of two wavelengths. The diffraction pattern is focused on a screen using a convex lens of focal length 50 cm placed near the slit. It is observed that the 1^ st secondary maximum of the first wavelength ₁ = 600 nm coincides exactly with the 2^ nd minimum of the unknown second wavelength ₂ . The linear width of the central maximum for the second wavelength ₂ on

Step-by-step solution

The position of the 1^ st secondary maximum for wavelength ₁ from the center is given by: y₁ = 3 ₁ f 2a The position of the 2^ nd minimum for wavelength ₂ from the center is given by: y₂ = 2 ₂ f a Given that these two positions coincide ( y₁ = y₂ ): 3 ₁ f 2a = 2 ₂ f a 3 ₁ 2 = 2 ₂ ₂ = 3 4 ₁ Substituting ₁ = 600 nm : ₂ = 3 4 600 nm = 450 nm The linear width of the central maximum for wavelength ₂ is: W = 2 ₂ f a Given f = 50 cm = 0.5 m and a = 0.1 mm = 10⁻⁴ m : W = 2 (450 10⁻⁹) 0.5 10⁻⁴ W = 450 10⁻⁹ 10⁻⁴ = 450 10⁻⁵ m