A parallel beam of light of wavelength 4000 A ∘ passes through a slit of width 5 × 10 - 3 m . The angular spread of the central maxima in the diffraction pattern is

Angular spread of the central maxima in the diffraction pattern = 2 λ e ( e = width of the slit) Here, e = 5 × 10 - 3 m , λ = 4000 Å Hence, 2 θ = 2 4000 × 10 - 10 5 × 10 - 3 = 1600 × 10 - 7 rad or 2 θ = 1.6 × 10 - 4 rad

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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsDiffractionEasy2 minPYQ_2020

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A parallel beam of light of wavelength $4000 \overset{\circ}{A}$ passes through a slit of width $5 \times 10^{- 3} m .$ The angular spread of the central maxima in the diffraction pattern is

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Solution:

Angular spread of the central maxima in the diffraction pattern $= \frac{2 λ}{e}$ ( $e$ = width of the slit)

Here, $e = 5 \times 10^{- 3} m, λ = 4000 Å$

Hence, $2 θ = 2 (\frac{4000 \times 10^{- 10}}{5 \times 10^{- 3}})$

$= 1600 \times 10^{- 7}$ rad

or $2 θ = 1.6 \times 10^{- 4}$ rad

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Wave OpticsSubtopic:Diffraction

⏱ 2mℹ️ Source: PYQ_2020

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