JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Geometrical Optics

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Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :

(A)9.52 cm
(B)18.23 cm
(C)37.58 cm✓ Correct
(D)55.44 cm

Answer: (C)

Solution

Light passes from region

I

into region

II

, so

n_{1} = 1.25

n_{2} = 1.40

and the object distance is

u = - 40

cm. The centre of curvature lies on the same side as the object, so

R = - 25

cm. Refraction at a single spherical surface gives

\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

\frac{1.40}{v} + \frac{1.25}{40} = \frac{1.40 - 1.25}{- 25} = - 0.006

\frac{1.40}{v} = - 0.006 - 0.03125 = - 0.03725

and

v = - 37.58

cm. The negative sign places the image on the same side as the object, at a distance of

37.58

cm from the surface.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Geometrical Optics

Solution

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