Question

On the axis of a transparent sphere of refractive index (n = 2) & radius 8 cm, an object is kept at the distance 8 cm from the surface of the sphere. Find the minimum distance of the image after all possible refraction from the surface of the sphere in cm.

Answer 1

8

Answer 2

The object is placed on the axis of a transparent sphere of refractive index $n = 2$ and radius $R = 8$ cm, at a distance of 8 cm from the surface of the sphere. Let the object be to the left of the sphere. The first refracting surface is the left surface of the sphere. The object distance from this surface is $u_1 = -8$ cm. The light is going from air ($n_1 = 1$) to the sphere ($n_2 = 2$). The radius of curvature of the left surface is $R_1 = +R = +8$ cm. Using the refraction formula at a spherical surface:

$\frac{n_2}{v_1} - \frac{n_1}{u_1} = \frac{n_2 - n_1}{R_1}$

$\frac{2}{v_1} - \frac{1}{-8} = \frac{2 - 1}{8}$

$\frac{2}{v_1} + \frac{1}{8} = \frac{1}{8}$

$\frac{2}{v_1} = 0$

$v_1 = \infty$

The image formed after the first refraction is at infinity. This means that the rays inside the sphere are parallel to the axis.

These parallel rays inside the sphere are incident on the right surface of the sphere. The light is going from the sphere ($n_2 = 2$) to air ($n_1 = 1$). The object for the second refraction is at $u_2 = \infty$. The radius of curvature of the right surface is $R_2 = -R = -8$ cm. Using the refraction formula at the right spherical surface:

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

$\frac{1}{v_2} - \frac{2}{\infty} = \frac{1 - 2}{-8}$

$\frac{1}{v_2} - 0 = \frac{-1}{-8} = \frac{1}{8}$

$\frac{1}{v_2} = \frac{1}{8}$

$v_2 = 8$ cm.

The final image is formed at a distance of 8 cm from the right surface of the sphere. Since the right surface is at a distance $R = 8$ cm from the center, and the image is formed 8 cm to the right of the right surface, the distance of the image from the center is $8 + 8 = 16$ cm.

The question asks for the minimum distance of the image after all possible refraction from the surface of the sphere in cm. The image is formed on the axis at a distance of 8 cm from the right surface. The right surface is part of the sphere. The minimum distance from a point on the axis to the surface of the sphere is the distance to the point where the axis intersects the surface. In this case, the image is at a distance of 8 cm from the right surface. This is the distance measured along the axis. Since the image is on the axis, the minimum distance from the image to the surface of the sphere is the perpendicular distance from the image point to the tangent plane at the point on the surface closest to the image. However, in the context of distances from the surface on the axis, it usually refers to the distance measured along the axis from the image point to the point where the axis intersects the surface.

Thus, the distance of the image from the right surface is 8 cm. This is a possible distance of the image from the surface. Since the image is formed outside the sphere, the distance from the surface is 8 cm.