A beam of light of wavelength 600 nm is incident on a single slit of width 0.2 mm. The distance to the screen is 2 m. Calculate the width of the central maximum.

B. 12 mm — The width of the central maximum in a single slit diffraction pattern is given by 2 * lambda * D / a. Substituting the given values, we get 2 * 600 * 10^-9 * 2 / 0.2 * 10^-3 = 12 * 10^-3 m.

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$A beam of light of wavelength λ = 600 nm is incident on a single slit of width a = 0.2 mm . The distance to the screen is D = 2 m . Calculate the width of the central maximum.$

A. $6 mm$
B. $12 mm$
C. $3 mm$
D. $24 mm$

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Correct answer: B — $12 mm$

Solution

The width of the central maximum is w = \frac{2 λ D}{a} . Substituting the values, w = \frac{2 \times 600 \times 1 0 ^{- 9} \times 2}{0.2 \times 1 0 ^{- 3}} = \frac{2400 \times 1 0 ^{- 9}}{0.2 \times 1 0 ^{- 3}} = 12 \times 1 0^{- 3} m = 12 mm .

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A beam of light of wavelength λ=600 nm is incident on a single slit of width a=0.2 mm. The distance to the screen is D=2 m. Calculate the width of the central maximum.

Solution

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More Wave Optics practice questions

$A beam of light of wavelength λ = 600 nm is incident on a single slit of width a = 0.2 mm . The distance to the screen is D = 2 m . Calculate the width of the central maximum.$