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JEE Advanced Physics Ray Optics 2024 JEE Advanced 2024 (Paper 1)

JEE Advanced Physics Question (2024) — Solution

Question

A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence _0. The ray partially refracts into the sphere with angle of refraction _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 . Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

Options

  1. A. P 5 ; Q 2 ; R 1 ; S 4
  2. B. P 5 ; Q 1 ; R 2 ; S 4
  3. C. P 3 ; Q 2 ; R 1 ; S 4
  4. D. P 3 ; Q 1 ; R 2 ; S 5

Answer

A. P 5 ; Q 2 ; R 1 ; S 4

Step-by-step solution

aligned & = ( _0- _0 )+ (180-2 _0 )+ ( _0-2 _0 ) \\ & =180+2 _0-4 _0 aligned (P) =180+2 _0-4 _0 180=180+2 _0-4 _0 _0=2 _0 ...(i) _0=2 _0 ...(ii) From (i) & (ii) aligned & _0=2 ( _0 / 2 ) ( _0 2 )=1 \\ & _0 2 =0 \\ & _0=0 aligned (Q) _0=2 _0 ...(i) _0= 3 _0 ...(ii) From (i) & (ii) aligned & _0= 3 ( _0 2 ) \\ & ( _0 2 )= 3 2 \\ & _0 2 =30,150 \\ & _0=60,300 (Rejected) \\ & _0=60 , 0 aligned (R) _0=2 _0 aligned & _0= 3 _0 \\ & 2 _0= 3 _0 \\ & _0= 3 2 \\ & _0=30,150 (Rejected) aligned _0=30, 0 ...(iii) aligned & (S) 45= 2 _0 \\ & _0=1 / 2 \\ & _0=60 \\ & =180+2 _0-4 _0 aligned =180+90-120 ...(iv) =180-30 ; =150^

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