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Question:

Two identical glass rods S1 and S2 (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 S1 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 S2. The distance d is:

80 cm

70 cm

90 cm

60 cm

Solution:

For S1, μ₂/v - μ₁/u = (μ₂ - μ₁)/R
Substituting the respective values we get,
1/v - 1.5/(-50) = (1 - 1.5)/10
or 1/v + 3/100 = -1/20
v = -50 cm
This acts as an image for S2 with distance from its aperture as d - 50 cm. For parallel rays to emerge from S2, the image must be at infinity. Therefore, for S2:
μ₂/v - μ₁/u = (μ₂ - μ₁)/R
1/∞ - 1.5/(d-50) = (1 - 1.5)/10
0 - 1.5/(d - 50) = -0.5/10
1.5/(d - 50) = 1/20
d - 50 = 30
d = 80 cm