A small transparent slab containing material of µ = 1.5 is placed along AS₂(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

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Vishal Baghel | Contributor-Level 10

7 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as the refractive index of the class $μ$ , the path difference will be calculated as $? x$ =2dsin $θ$ +( $μ - 1$ )L

For principal maxima ,(path difference is zero)

2dsin $θ$ ₀+( $μ - 1$ )L=0

Sin $θ$ ₀= - $\frac{L (μ - 1)}{2 d}$ = $\frac{- L (0.5)}{2 d}$

Sin $θ$ ₀=-1/16

OP=Dtan $θ$ ₀= Dsin $θ$ ₀=-D/16

For p

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3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\Rightarrow \frac{d y}{D} = (2 n + 1) \frac{λ}{2}$ [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{d y}{D} = \frac{λ}{2}$

$λ = \frac{2 d}{D} y$

$= \frac{2 d}{D} \times \frac{d}{2} [y = \frac{d}{2}, G i v e n$ first dark fringe is observed on the screen directly opposite to one of the slits]

$λ = \frac{2 \times 0.6 \times 0.6}{800 \times 2}$

$λ = 450 m m$

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A small transparent slab containing material of µ = 1.5 is placed along AS2(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?