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Standard XII

Physics

Introduction to Wave Optics

A monochromat...

Question

A monochromatic light is traveling in a medium of refractive index $n = 1.6$ . It enters a stack of glass layer from the bottom side at an angle $θ = 30^{\circ} .$ The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layer are monotonically decreasing as $n_{m} = n - m Δ n$ , where $n_{m}$ is the refractive index of the $m^{t h}$ slab and $Δ n = 0.1$ (see the figure). The ray is refracted out parallel to the interface between the $(m - 1)^{t h}$ and $m^{t h}$ slabs from the right side of the stack. What is the value of $m$ is

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Solution

Applying Snell's law,
$n sin θ = n_{m} sin 90^{\circ}$
$n sin 30^{\circ} = n_{m} sin 90^{\circ}$

$n \times \frac{1}{2} = (n - m Δ n) (1)$

$\frac{1.6}{2} = [1.6 - m (0.1)]$

$m (0.1) = \frac{1.6}{2}$

$m = 8$

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Applying Snell's law, nsin θ=nmsin90∘ nsin30∘=nmsin90∘ n×12=(n−mΔn)(1) 1.62=[1.6−m(0.1)] m(0.1)=1.62 m=8

Applying Snell's law,
$n sin θ = n_{m} sin 90^{\circ}$
$n sin 30^{\circ} = n_{m} sin 90^{\circ}$

$n \times \frac{1}{2} = (n - m Δ n) (1)$

$\frac{1.6}{2} = [1.6 - m (0.1)]$

$m (0.1) = \frac{1.6}{2}$

$m = 8$