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Question

Physics Question on Ray optics and optical instruments

Two identical glass rods S1S_1 and S2S_2 (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S1S_1 on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S2S_2. The distance d is

A

60 cm

B

70 cm

C

80 cm

D

90 cm

Answer

70 cm

Explanation

Solution

This question is of multiple event. First event on curved surface of S1S_1 & Second on curved surface of S2S_2 Event 1n2=1- 1\quad n_{2} = 1 n1=1.5n_{1}= 1.5 1v1.5u=11.5R\frac{1}{v} - \frac{1.5}{u} = \frac{1-1.5}{R} 1v1.550=0.510\frac{1}{v}- \frac{1.5}{-50} = \frac{-0.5}{-10} 1v+1.550=120\frac{1}{v} + \frac{1.5}{50} = \frac{1}{20} 1v=1201.550\frac{1}{v} = \frac{1}{20}-\frac{1.5}{50} 1v=53100=2100?v=50cm\frac{1}{v} = \frac{5-3}{100} = \frac{2}{100}\quad\quad? v = 50 \,cm for second event u=(d50)\quad u = - \left(d - 50\right) v=8v = 8 R=+10cmR = + 10\, cm 1.5v1u=1.51R\frac{1.5}{v}-\frac{1}{u} = \frac{1.5-1}{R} 1.51(d50)=0.510\frac{1.5}{\infty} - \frac{1}{-\left(d-50\right)} = \frac{0.5}{10} 1d50=120\frac{1}{d-50} = \frac{1}{20} 20=d5020 = d - 50 d=70cmd = 70 \,cm