Question

A point object O is placed in front of a glass rod having spherical end of radius of curvature 30 cm. The image formed would be

• A. 30 cm left
• B. infinity
• C. 1 cm to the right
• D. 18 cm to the left

Answer 1

Answer: (A)

30 cm left

Answer 2

The problem involves refraction at a single spherical surface. We are given a point object placed in front of a glass rod with a spherical end.

Given information:

• Object distance, $u = -15$ cm (The object is to the left of the spherical surface, and we use the sign convention where light travels from left to right).
• Radius of curvature of the spherical surface, $R = +30$ cm (The surface is convex towards the object, so the center of curvature is to the right of the surface).
• Refractive index of the first medium (air), $n_1 = 1$.
• Refractive index of the second medium (glass), $n_2 = 1.5$ (This is a standard assumption for glass).

Formula for Refraction at a Spherical Surface: The formula relating the object distance ($u$), image distance ($v$), refractive indices of the two media ($n_1$ and $n_2$), and the radius of curvature ($R$) is: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$

Applying the formula: Substitute the given values into the formula: $\frac{1.5}{v} - \frac{1}{-15 \text{ cm}} = \frac{1.5 - 1}{+30 \text{ cm}}$ $\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$

Now, isolate the term with $v$: $\frac{1.5}{v} = \frac{0.5}{30} - \frac{1}{15}$ To subtract the fractions on the right side, find a common denominator, which is 30: $\frac{1.5}{v} = \frac{0.5}{30} - \frac{1 \times 2}{15 \times 2}$ $\frac{1.5}{v} = \frac{0.5}{30} - \frac{2}{30}$ $\frac{1.5}{v} = \frac{0.5 - 2}{30}$ $\frac{1.5}{v} = \frac{-1.5}{30}$

Solve for $v$: $v = \frac{1.5 \times 30}{-1.5}$ $v = -30 \text{ cm}$

The negative sign for $v$ indicates that the image is formed on the same side of the spherical surface as the object. Therefore, the image is formed 30 cm to the left of the spherical surface.