Consider single slit experiment of diffraction of light. If the light of wavelength 5000 A ∘ fall on a slit of width 1 μ m then the angular width of central maximum is

d sin ⁡ θ = λ sin ⁡ θ = λ d For first minima = 5 × 10 - 7 10 - 6 sin ⁡ θ = 5 10 θ = 30 ° Angular width of first minima 2 θ = 60 °

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PhysicsWave OpticsDiffractionEasy2 minQB

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Consider single slit experiment of diffraction of light. If the light of wavelength $5000 \overset{\circ}{A}$ fall on a slit of width $1 μ m$ then the angular width of central maximum is

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Answer:

Solution:

$d \sin θ = λ$

$\sin θ = \frac{λ}{d}$

For first minima

$= \frac{5 \times 10^{- 7}}{10^{- 6}}$

$\sin θ = \frac{5}{10}$

$θ = 30 °$ Angular width of first minima

$2 θ = 60 °$

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:Diffraction

⏱ 2mℹ️ Source: QB

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