Question

. A transparent sphere of radius ‘y’ and refractive index n is kept in air. At what distance from the surface of the sphere should a point object be placed on the principal axis so as to form a real image at the same distance from the second surface of the sphere?

Answer 1

Answer:

The object should be placed at the distance of \frac{R}{\mu - 1}

μ−1

R

from the surface of the sphere.

Explanation:

Radius of sphere = R

Refractive Index of the sphere = \muμ

Using Equation

\frac{\mu_1}{v} - \frac{\mu_2}{u} = \frac{\mu_2-\mu_1}{R}

v

μ

1

−

u

μ

2

=

R

μ

2

−μ

1

For the refraction at the first surface of the sphere.(Air to glass)

\frac{\mu}{\infty}+\frac{1}{x} = \frac{\mu - 1}{R}

∞

μ

+

x

1

=

R

μ−1

[v = ∞, \mu_2 = \muμ

2

=μ , u = -x, \mu_1μ

1

= 1]

\frac{\mu}{\infty} + \frac{1}{x} = \frac{\mu - 1}{R}

∞

μ

+

x

1

=

R

μ−1

⇒\frac{1}{x} = \frac{\mu - 1}{R}

x

1

=

R

μ−1

⇒x = \frac{R}{\mu - 1}x=

μ−1

R

Hence,the object should be placed at the distance of \frac{R}{\mu - 1}

μ−1

R

from the surface of the sphere to obtain the real image at the distance from the second surface of the sphere.