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Geometrical Optics

Section: Physics

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Question : 38 of 59

Marks: +1, -0

Two identical glass rods

S_1 \text{ and } S_2

(refractive index=1.5) have one convex end of radius of curvature 10 cm.They are planed with the curved surfaces at a distance d as a shown in the figure figure.with their axes (shown by the dashed line)aligned.when a point source of light P is placed inside rod

S_1

on its axis at a distance of 50 cm from the curved face ,the light rays emanating from it are found to br parallel t o the axis inside

S_2

.The distance d is

[JEE Adv 2015 P1]

60 cm
70 cm
80 cm
90 cm

Solution:

For 1st refraction

\frac{1}{v} - \frac{1.5}{-50} = \frac{1-1.5}{-10}

\Rightarrow v = 50 \text{ cm}

For 2nd refraction

\frac{1.5}{\infty} - \frac{1}{-x} = \frac{1.5-1}{-10}

\Rightarrow x = 20 \text{ cm}

\Rightarrow d = 70 \text{ cm}

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