Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is 312.5 × 10 - 3 c m . If waves were incident normally on the slit, then wavelength of the light from the source will be

Width of central max. = 2 λ D a λ = 0.2 × 10 - 2 × 312.5 × 10 - 3 × 10 - 2 2 × 1 2 m = 6250 × 10 - 10 m = 6250 Å

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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsDiffractionEasy2 minQB

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Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is $312.5 \times 10^{- 3} c m$ . If waves were incident normally on the slit, then wavelength of the light from the source will be

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Solution:

Width of central max. $= \frac{2 λ D}{a}$

$λ = \frac{0.2 \times 10^{- 2} \times 312.5 \times 10^{- 3} \times 10^{- 2}}{2 \times \frac{1}{2}} m$

$= 6250 \times 10^{- 10} m = 6250 Å$

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:Diffraction

⏱ 2mℹ️ Source: QB

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