A light ray is incident on the surface of a sphere of refractive index at an angle of incidence . The ray partially refracts into the sphere with angle of refraction and then partly reflects from the back surface. [2024]

Question

A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $θ_{0}$ . The ray partially refracts into the sphere with angle of refraction $ϕ_{0}$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $α$ . Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [2024]

	List - I		List - II
(P)	If $n = 2$ and $α = 180 °$ , then all the possible values of $θ_{0}$ will be	(1)	$30 °$ and $0 °$
(Q)	If $n = \sqrt{3}$ and $α = 180 °$ , then all the possible values of $θ_{0}$ will be	(2)	$60 °$ and $0 °$
(R)	If $n = \sqrt{3}$ and $α = 180 °$ , then all the possible values of $ϕ_{0}$ will be	(3)	$45 °$ and $0 °$
(S)	If $n = \sqrt{2}$ and $θ_{0} = 45 °$ , then all the possible values of $α$ will be	(4)	$150 °$
		(5)	$0 °$

Smart Achievers · Accepted Answer

(1)

Total angle of deviation of the emergent ray with respect to the incident ray,

$α = (θ_{0} - ϕ_{0}) + (π - 2 ϕ_{0}) + (θ_{0} - ϕ_{0})$

$α = π + 2 θ_{0} - 4 ϕ_{0}$ $\dots (i)$

$\sin θ_{0} = n \sin ϕ_{0}$ $\dots (i i)$

For $(P)$ , $n = 2, α = 180 °$

If $α = π$ , $2 θ_{0} - 4 ϕ_{0} = 0$

$θ_{0} = 2 ϕ_{0}$

$\sin θ_{0} = 2 \sin (\frac{θ_{0}}{2})$

For $(Q)$ , $n = \sqrt{3}, α = 180 °$

$θ_{0} = 2 ϕ_{0}$

$\sin θ_{0} = \sqrt{3} \sin (\frac{θ_{0}}{2})$

or, $θ_{0} = 60 °, 0 °$

For $(R)$ , $n = \sqrt{3}, α = 180 °$

$θ_{0} = 2 ϕ_{0}$

$\sin θ_{0} = \sqrt{3} \sin ϕ_{0}$

$\sin 2 θ_{0} = \sqrt{3} \sin ϕ_{0}$

$\cos ϕ_{0} = \frac{\sqrt{3}}{2}$

$∴$ $ϕ_{0} = 30 °, 0 °$

For $(S)$ , $n = \sqrt{2}, θ_{0} = 45 °$

$\sin 45 ° = \sqrt{2} \cos ϕ_{0}$

$\cos ϕ_{0} = \frac{1}{2}$ or $ϕ_{0} = 60 °$

$α = 180 ° + 2 θ_{0} - 4 ϕ_{0}$

or, $α - 180 ° + 90 ° - 120 ° = 150 °$

Solution

1 P → 5; Q → 2; R → 1; S → 4

2 P → 5; Q → 1; R → 2; S → 4

3 P → 3; Q → 2; R → 1; S → 4

4 P → 3; Q → 1; R → 2; S → 5

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