JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Geometrical Optics

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A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is ____ mm.

Answer: 225

Solution

For refraction at the first surface (air to glass), using

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

with

μ_{1} = 1

μ_{2} = 1.5

u = \infty

R = 15

cm:

\frac{1.5}{v} = \frac{0.5}{15} ⟹ v = 45 cm

This image acts as object for the second surface (glass to air) at distance

u^{'} = 45 - 30 = 15

cm from the second surface.

For the second surface with

R = - 15

cm:

\frac{1}{v ^{'}} - \frac{1.5}{15} = \frac{1 - 1.5}{- 15}

\frac{1}{v ^{'}} = \frac{1}{10} + \frac{1}{30} = \frac{4}{30}

v^{'} = 7.5 cm from second surface

Distance from center =

15 + 7.5 = 22.5

cm =

225

mm.

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Geometrical Optics

Solution

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