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QJEEAdv15Ph4db0: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph4db0 Solution: Two identical glass rods S 1 and S 2 (refractive index = 1.5)…

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
JEE Advanced 2015 Paper 1 Offline Physics - Geometrical Optics Question 38 English

Multiple Choice Options

  • A 60 cm
  • B 70 cm
  • C 80 cm
  • D 90 cm

Step-by-Step Verified Solution

Apply the formula for the refraction at the spherical surface of S1, μ2v−μ1u=μ2−μ1R{{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{\mu _2} - {\mu _1}} \over R} , to get

1.0v−1.5(−50)=1−1.5(−10){{1.0} \over v} - {{1.5} \over {( - 50)}} = {{1 - 1.5} \over {( - 10)}} .

Solve to get the image distance v = 50 cm (point Q in the figure).

JEE Advanced 2015 Paper 1 Offline Physics - Geometrical Optics Question 38 English Explanation

The image Q acts as an object for refraction at the spherical surface of S2. The object distance is u=−(d−50)u = - (d - 50) . The image distance is v = ∞\infty (because rays are parallel in S2). Apply the formula for refraction at the spherical surface of S2, to get

1.5∞−1−(d−50)=1.5−1(10){{1.5} \over \infty } - {1 \over { - (d - 50)}} = {{1.5 - 1} \over {(10)}} .

Solve to get d = 70 cm.

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