Question

A small object is placed at distance x = 20 cm from of a glass sphere of radius R such that its real image is having magnification 1. If refractive index of glass is 1.5, then R is

Answer 1

10 cm

Answer 2

For a glass sphere, if an object is placed at a distance 'x' from its first surface such that a real image of magnification -1 is formed, then the image is also formed at a distance 'x' from the second surface. This specific configuration occurs when the object is placed at the first principal focal point of the sphere, measured from the surface, which is given by $x = \frac{R}{n-1}$, where R is the radius of the sphere and n is its refractive index.

Given:

• Object distance $x = 20$ cm.
• Refractive index of glass $n = 1.5$.

Using the formula: $x = \frac{R}{n-1}$

$20 = \frac{R}{1.5 - 1}$

$20 = \frac{R}{0.5}$

$R = 20 \times 0.5$

$R = 10$ cm.

This result can be verified by applying the spherical refraction formula for both surfaces:

1. For the first surface (air to glass): $\frac{n_2}{v_1} - \frac{n_1}{u_1} = \frac{n_2 - n_1}{R_1}$

   $\frac{1.5}{v_1} - \frac{1}{-20} = \frac{1.5 - 1}{R}$

   $\frac{1.5}{v_1} + \frac{1}{20} = \frac{0.5}{R}$

   If $R = 10$ cm, then $\frac{1.5}{v_1} + \frac{1}{20} = \frac{0.5}{10} = \frac{1}{20}$.

   This implies $\frac{1.5}{v_1} = 0$, so $v_1 = \infty$. This means rays become parallel after the first refraction.

2. For the second surface (glass to air): The parallel rays (object at infinity) fall on the second surface.

   $\frac{n_2'}{v_2} - \frac{n_1'}{u_2} = \frac{n_2' - n_1'}{R_2}$

   Here, $n_1' = 1.5$, $n_2' = 1$, $u_2 = \infty$, $R_2 = -R = -10$ cm.

   $\frac{1}{v_2} - \frac{1.5}{\infty} = \frac{1 - 1.5}{-10}$

   $\frac{1}{v_2} = \frac{-0.5}{-10} = \frac{0.5}{10} = \frac{1}{20}$.

   $v_2 = 20$ cm.

This confirms that the image is formed at 20 cm from the second surface, which is consistent with the condition for magnification -1.