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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsYDSEEasy2 minPYQ_2020

PhysicsEasysingle choice

When a thin transparent sheet of refractive index $μ = \frac{3}{2}$ is placed near one of the slit in YDSE, the intensity at centre of screen reduces to half of maximum intensity. Then, minimum thickness of sheet should be

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Answer:

Solution:

$l_{R} = 4 l_{0} \cos^{2} (\frac{Δ θ}{2})$

As given $l_{R}$ at centre is $2 l_{0}$ , then $2 l_{0} = 4 l_{0} \cos^{2} (\frac{Δ θ}{2})$

$\Rightarrow$ $(\frac{Δ θ}{2}) = \frac{π}{4}$ [for minimum result]

$Δ θ = \frac{π}{2}$

$Δ x = Δ θ \times \frac{λ}{2 π} = \frac{λ}{4}$

For slab $Δ x = (μ - 1) t$

$(\frac{3}{2} - 1) t = \frac{λ}{4}$

$t = \frac{λ}{2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Wave OpticsSubtopic:YDSE

⏱ 2mℹ️ Source: PYQ_2020

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