Question

A glass spherical paper weight of refractive index $\frac{3}{2}$ and radius 15 cm is placed on a page of a book (choose a letter of the print vertically below the centre of flat base). The position of image of the print from top of sphere is

• A. 60 cm below
• B. 60 cm above
• C. 30 cm below
• D. 15 cm below

Answer 1

Answer: (D)

15 cm below

Answer 2

Let $n_1 = 1$ be the refractive index of air and $n_2 = \frac{3}{2}$ be the refractive index of the glass. The radius of the spherical paperweight is $R = 15$ cm. The paperweight is placed on a book, and the print is vertically below the center of the flat base. This means the print is at the center of the sphere. We consider the refraction at the lower spherical surface of the paperweight. Light travels from air ($n_1$) into glass ($n_2$). The object is the print, which is at the center of the sphere. Let's set the origin at the center of the sphere. The print is at the origin ($0$). The vertex of the lower spherical surface is at a distance $R$ below the center, i.e., at $-R$. The object distance $u$ is measured from the vertex of the refracting surface. Thus, $u = 0 - (-R) = R$. So, the object distance is $u = R = 15$ cm.

We use the formula for refraction at a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_{curvature}}$ In this case, light is entering the glass sphere from air. The refracting surface is the lower hemisphere. The center of curvature of this surface is the center of the sphere. Since the object is inside the sphere and the light is entering from below, the center of curvature is on the right side of the vertex (if we consider the vertex at $-R$ and the center at $0$). Thus, the radius of curvature $R_{curvature}$ is positive. $n_1 = 1$, $n_2 = \frac{3}{2}$, $u = 15$ cm, $R_{curvature} = 15$ cm.

Substituting these values into the formula: $\frac{3/2}{v} - \frac{1}{15} = \frac{3/2 - 1}{15}$ $\frac{3/2}{v} - \frac{1}{15} = \frac{1/2}{15}$ $\frac{3/2}{v} = \frac{1}{15} + \frac{1}{30}$ $\frac{3/2}{v} = \frac{2+1}{30} = \frac{3}{30} = \frac{1}{10}$ $v = \frac{3}{2} \times 10 = 15 \text{ cm}$ The image distance $v$ is measured from the vertex of the lower spherical surface. The vertex is at $-R$ (i.e., $-15$ cm from the center). The position of the image from the center is $v_{center} = -R + v = -15 \text{ cm} + 15 \text{ cm} = 0$. So, the image is formed at the center of the sphere.

The question asks for the position of the image from the top of the sphere. The top of the sphere is at a distance $R$ above the center. The position of the image is at the center. The position of the top of the sphere is at $R$ from the center. Therefore, the distance of the image from the top of the sphere is $R - 0 = R = 15$ cm. Since the image is at the center and the top is above the center, the image is below the top of the sphere.

The position of the image from the top of the sphere is 15 cm below.