Question

A glass sphere ($\mu=1.5$) of radius 12 $cm$ is placed in sunlight. Where is the image of the sun formed by the light passing through the sphere?

• A. 8 $cm$ from second surface
• B. 6 $cm$ from second surface
• C. 4 $cm$ from second surface
• D. 12 $cm$ from second surface

Answer 1

Answer: (B)

6 cm from second surface

Answer 2

The problem involves finding the image formed by a thick spherical lens (a glass sphere) for an object at infinity (the sun). This requires applying the spherical refraction formula twice, once for each surface of the sphere.

Given:

• Refractive index of the glass sphere ($\mu_g$) = 1.5
• Radius of the sphere (R) = 12 cm
• Object (sun) is at infinity, so $u_1 = -\infty$.
• Refractive index of air ($\mu_a$) = 1

Step 1: Refraction at the first surface

Light from the sun enters the sphere from air to glass. The formula for refraction at a spherical surface is:

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

For the first surface:

• $\mu_1 = \mu_a = 1$ (air)
• $\mu_2 = \mu_g = 1.5$ (glass)
• $u_1 = -\infty$ (object at infinity)
• $R_1 = +12$ cm (The first surface is convex for incident light from the left, so its center of curvature is to the right).

Substituting these values into the formula:

$\frac{1.5}{v_1} - \frac{1}{-\infty} = \frac{1.5 - 1}{+12}$

$\frac{1.5}{v_1} - 0 = \frac{0.5}{12}$

$\frac{1.5}{v_1} = \frac{0.5}{12}$

$v_1 = \frac{1.5 \times 12}{0.5} = 3 \times 12 = 36$ cm

The first image ($I_1$) is formed at 36 cm to the right of the first surface, inside the sphere.

Step 2: Refraction at the second surface

The first image ($I_1$) acts as the object for the second surface. Light travels from glass to air. The diameter of the sphere is $2R = 2 \times 12 = 24$ cm. The first image $I_1$ is formed at 36 cm from the first surface. Therefore, the distance of $I_1$ from the second surface is $u_2 = v_1 - 2R = 36 \text{ cm} - 24 \text{ cm} = 12$ cm. Since $I_1$ is to the right of the second surface, and light is incident from the left (inside the sphere), this is a virtual object, so $u_2 = +12$ cm.

For the second surface:

• $\mu_1' = \mu_g = 1.5$ (glass)
• $\mu_2' = \mu_a = 1$ (air)
• $u_2 = +12$ cm (virtual object)
• $R_2 = -12$ cm (The second surface is concave for light coming from inside the sphere, so its center of curvature is to the left).

Substituting these values into the formula:

$\frac{1}{v_2} - \frac{1.5}{+12} = \frac{1 - 1.5}{-12}$

$\frac{1}{v_2} - \frac{1.5}{12} = \frac{-0.5}{-12}$

$\frac{1}{v_2} - \frac{1.5}{12} = \frac{0.5}{12}$

$\frac{1}{v_2} = \frac{0.5}{12} + \frac{1.5}{12}$

$\frac{1}{v_2} = \frac{0.5 + 1.5}{12}$

$\frac{1}{v_2} = \frac{2.0}{12}$

$\frac{1}{v_2} = \frac{1}{6}$

$v_2 = +6$ cm

The final image is formed at 6 cm to the right of the second surface, in the air.