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प्रश्न
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)?
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उत्तर
Data: 2W = 6 mm ∴ W = 3 mm = 3 × 10-3 m, y = 2.5 m,
(a) λ1 = 500 nm = 5 × 10-7 m
(b) λ2 = 50 µm = 5 × 10-5 m
(c) λ3 = 0.500 nm = 5 × 10-10 m
Let a be the slit width.
(a) W = `("y" lambda_1)/"a"`
∴ a = `("y" lambda_1)/"W" = ((2.5)(5 xx 10^-7))/(3 xx 10^-3)`
= 4.167 × 10-4 m
= 0.4167 mm
(b) W = `("y" lambda_2)/"a"`
∴ a = `("y" lambda_2)/"W" = ((2.5)(5 xx 10^-5))/(3 xx 10^-3)`
= 4.167 × 10-2 m
= 41.67 mm
(c) W = `("y" lambda_3)/"a"`
∴ a = `("y" lambda_3)/"W" = ((2.5)(5 xx 10^-10))/(3 xx 10^-3)`
= 4.167 × 10-7 m
= 4.167 × 10-4 mm
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