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

The problem involves refraction at a spherical surface. We assume the print is located at the center of the spherical paperweight. Given: Refractive index of glass, $n_1 = \frac{3}{2}$ Refractive index of air, $n_2 = 1$ Radius of the sphere, $R = 15$ cm

We use the formula for refraction at a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_{curvature}}$ Here, the object is at the center of the sphere. The pole of the bottom surface is at the bottom of the sphere. The object distance from the pole is $u = R = 15$ cm. The object is in air ($n_1 = 1$), and the image is formed in glass ($n_2 = \frac{3}{2}$). The radius of curvature for the bottom surface is $R_{curvature} = +R = +15$ cm (since the center of curvature is on the side where the image is formed).

Substituting the values: $\frac{\frac{3}{2}}{v} - \frac{1}{15} = \frac{\frac{3}{2} - 1}{15}$ $\frac{3}{2v} - \frac{1}{15} = \frac{\frac{1}{2}}{15}$ $\frac{3}{2v} - \frac{1}{15} = \frac{1}{30}$ $\frac{3}{2v} = \frac{1}{15} + \frac{1}{30} = \frac{2+1}{30} = \frac{3}{30} = \frac{1}{10}$ $v = \frac{3 \times 10}{2} = 15 \text{ cm}$ The image distance $v = 15$ cm is measured from the bottom pole. Since $v$ is positive, the image is formed 15 cm from the bottom pole, which is 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 $R = 15$ cm above the center. Therefore, the image, which is at the center, is 15 cm below the top of the sphere.