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PhysicsMultiple correct+4 / -2Question 14 of 60

JEE Advanced 2015 · Paper 1, Physics question 14

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 dd as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light PP 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 dd is

Two coaxial glass rods with convex ends facing each other across an air gap d. A point source P lies inside rod S₁ on the axis, 50 cm from its curved face; rod S₂ is on the right.
Choose one or more options
Answer and worked solution

Official answer: B

First surface (glass to air). n1=1.5n_1=1.5, n2=1n_2=1, u=−50u=-50 cm and R=−10R=-10 cm:
1v−1.5−50=1−1.5−10  ⟹  1v=0.05−0.03  ⟹  v=50 cm\frac1v-\frac{1.5}{-50}=\frac{1-1.5}{-10}\implies\frac1v=0.05-0.03\implies v=50\text{ cm}
into the air.

Second surface (air to glass). For parallel rays inside S2S_2 (v=∞v=\infty), with R=+10R=+10 cm:
−1u=0.510  ⟹  u=−20 cm.-\frac1u=\frac{0.5}{10}\implies u=-20\text{ cm}.

Gap. d=50+20=70d=50+20=70 cm.

(B)

Topics:Refraction at spherical surfaces
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