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JEE Advanced 2024 · Paper 1, Physics question 16

A light ray is incident on the surface of a sphere of refractive index nn at an angle of incidence θ0\theta_0. The ray partially refracts into the sphere with angle of refraction ϕ0\phi_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\alpha. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-I

  • P
    If n=2n = 2 and α=180∘\alpha = 180^{\circ}, then all the possible values of θ0\theta_0 will be
  • Q
    If n=3n = \sqrt{3} and α=180∘\alpha = 180^{\circ}, then all the possible values of θ0\theta_0 will be
  • R
    If n=3n = \sqrt{3} and α=180∘\alpha = 180^{\circ}, then all the possible values of ϕ0\phi_0 will be
  • S
    If n=2n = \sqrt{2} and θ0=45∘\theta_0 = 45^{\circ}, then all the possible values of α\alpha will be

List-II

  • 1
    30∘30^{\circ} and 0∘0^{\circ}
  • 2
    60∘60^{\circ} and 0∘0^{\circ}
  • 3
    45∘45^{\circ} and 0∘0^{\circ}
  • 4
    150∘150^{\circ}
  • 5
    0∘0^{\circ}
Choose one option
Answer and worked solution

Official answer: A

Deviation. Refraction at entry, one internal reflection, refraction at exit:
α=(θ0−ϕ0)+(180∘−2ϕ0)+(θ0−ϕ0)=180∘+2θ0−4ϕ0.\alpha = (\theta_0 - \phi_0) + (180^\circ - 2\phi_0) + (\theta_0 - \phi_0) = 180^\circ + 2\theta_0 - 4\phi_0.

When α=180∘\alpha = 180^\circ: θ0=2ϕ0\theta_0 = 2\phi_0. With Snell's law sin⁡θ0=nsin⁡ϕ0\sin\theta_0 = n\sin\phi_0: sin⁡2ϕ0=nsin⁡ϕ0\sin 2\phi_0 = n\sin\phi_0, so either ϕ0=0\phi_0 = 0 or cos⁡ϕ0=n2\cos\phi_0 = \frac n2.

  • (P) n=2n = 2: cos⁡ϕ0=1\cos\phi_0 = 1, so only θ0=0∘\theta_0 = 0^\circ. → (5)
  • (Q) n=3n = \sqrt3: ϕ0=30∘\phi_0 = 30^\circ, θ0=60∘\theta_0 = 60^\circ; or 0∘0^\circ. → (2)
  • (R) ϕ0=30∘\phi_0 = 30^\circ or 0∘0^\circ. → (1)
  • (S) n=2n = \sqrt2, θ0=45∘\theta_0 = 45^\circ: sin⁡ϕ0=12\sin\phi_0 = \frac12, ϕ0=30∘\phi_0 = 30^\circ, so α=180∘+90∘−120∘=150∘\alpha = 180^\circ + 90^\circ - 120^\circ = 150^\circ. → (4)

Option (A).

Topics:Refraction at plane surfacesReflection and plane mirrors
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