A monochromatic light is travelling in a medium of refractive index It enters a stack of glass layers from the bottom side at an angle The interfaces of the glass layers are parallel to each other. [2017]

Question

A monochromatic light is travelling in a medium of refractive index $n = 1.6 .$ It enters a stack of glass layers from the bottom side at an angle $θ = 30 ° .$ The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $n_{m} = n - m Δ n,$ where $n_{m}$ is the refractive index of the $m^{th}$ slab and $Δ n = 0.1$ (see the figure). The ray is refracted out parallel to the interface between the ${(m - 1)}^{th}$ and $m^{th}$ slabs from the right side of the stack. What is the value of $m$ [2017]

Smart Achievers · Accepted Answer

(8)

$Here, n \sin θ = (n - m Δ n) \sin 90 °$

$\Rightarrow n \times \sin 30 ° = (n - m \times 0.1) \sin 90 °$

$∴$ $1.6 \times \sin 30 ° = 1.6 - m \times 0.1$

$\Rightarrow 1.6 \times \frac{1}{2} = 1.6 - 0.1 m$

$0.8 = 1.6 - 0.1 m$

$0.1 m = 0.8$

$m = 8$

Solution

Ans.
(8)

$Here, n \sin θ = (n - m Δ n) \sin 90 °$

$\Rightarrow n \times \sin 30 ° = (n - m \times 0.1) \sin 90 °$

$∴$ $1.6 \times \sin 30 ° = 1.6 - m \times 0.1$

$\Rightarrow 1.6 \times \frac{1}{2} = 1.6 - 0.1 m$

$0.8 = 1.6 - 0.1 m$

$0.1 m = 0.8$

$m = 8$

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Solution

Ans. (8) Here, nsinθ=(n-mΔn)sin90° ⇒n×sin30°=(n-m×0.1)sin90° ∴ 1.6×sin30°=1.6-m×0.1 ⇒1.6×12=1.6-0.1m 0.8=1.6-0.1m 0.1m=0.8 m=8

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Ans.
(8)

$Here, n \sin θ = (n - m Δ n) \sin 90 °$

$\Rightarrow n \times \sin 30 ° = (n - m \times 0.1) \sin 90 °$

$∴$ $1.6 \times \sin 30 ° = 1.6 - m \times 0.1$

$\Rightarrow 1.6 \times \frac{1}{2} = 1.6 - 0.1 m$

$0.8 = 1.6 - 0.1 m$

$0.1 m = 0.8$

$m = 8$