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: (A)

60 cm below

Answer 2

The problem involves calculating the image position formed by refraction at a spherical surface. We consider the top surface of the spherical paperweight as the refracting interface.

Given: Refractive index of glass, $n_1 = \frac{3}{2}$ Refractive index of air, $n_2 = 1$ Radius of the sphere, $R = 15$ cm

The print is located vertically below the center of the flat base, which means it is at the bottom of the sphere. We are viewing the print from the top of the sphere.

Let's set up the coordinate system with the origin at the center of the sphere. The top of the sphere is at a distance $R$ from the center, and the bottom is at a distance $R$ from the center.

The object (print) is at the bottom of the sphere. The top surface of the sphere is where the refraction occurs as light exits the glass into the air.

The distance of the object from the top surface needs to be determined. The object is at the bottom of the sphere, and the top surface is at the top of the sphere. The total diameter of the sphere is $2R$. Therefore, the object distance from the top surface is $u = -2R$ (negative sign indicates the object is on the left of the refracting surface, assuming light travels from left to right).

The refracting surface is the top of the sphere. The center of curvature of this spherical surface is at the center of the sphere. Since the surface is convex to the incident light (light travels from the object inside the sphere towards the top surface), and the center of curvature is to the left of the surface, the radius of curvature is $R_{curv} = -R$.

We use the spherical refraction formula: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_{curv}}$

Substitute the given values: $n_1 = \frac{3}{2}$ $n_2 = 1$ $u = -2R$ $R_{curv} = -R$

$\frac{1}{v} - \frac{3/2}{-2R} = \frac{1 - 3/2}{-R}$ $\frac{1}{v} + \frac{3}{4R} = \frac{-1/2}{-R}$ $\frac{1}{v} + \frac{3}{4R} = \frac{1}{2R}$

Now, solve for $v$: $\frac{1}{v} = \frac{1}{2R} - \frac{3}{4R}$ $\frac{1}{v} = \frac{2}{4R} - \frac{3}{4R}$ $\frac{1}{v} = \frac{2 - 3}{4R}$ $\frac{1}{v} = -\frac{1}{4R}$ $v = -4R$

The image distance $v$ is measured from the top surface. The negative sign indicates that the image is formed on the same side as the object, which means the image is located below the top surface.

Substitute the value of $R = 15$ cm: $v = -4 \times 15 \text{ cm} = -60 \text{ cm}$

The position of the image is 60 cm below the top of the sphere.