JEE Advance - Physics (2015 - Paper 1 Offline - No. 14)
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
60 cm
70 cm
80 cm
90 cm
Explanation
Apply the formula for the refraction at the spherical surface of S1, $${{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{\mu _2} - {\mu _1}} \over R}$$, to get
$${{1.0} \over v} - {{1.5} \over {( - 50)}} = {{1 - 1.5} \over {( - 10)}}$$.
Solve to get the image distance v = 50 cm (point Q in the figure).
The image Q acts as an object for refraction at the spherical surface of S2. The object distance is $$u = - (d - 50)$$. 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
$${{1.5} \over \infty } - {1 \over { - (d - 50)}} = {{1.5 - 1} \over {(10)}}$$.
Solve to get d = 70 cm.
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