Question

Figure shows a glass hemisphere M of $\mu = 3/2$ and radius 10 cm. A point object O is placed at a distance 20 cm behind the flat face which is viewed by an observer from the curved side. Find location of final image after two refractions as seen by observer.

• A. 10 cm behind the flat face
• B. 10 cm in front of the flat face
• C. 30 cm behind the flat face
• D. 30 cm in front of the flat face

Answer 1

Answer: (A)

The final image is formed at a distance of 10 cm behind the flat face.

Answer 2

The problem involves two refractions: first at the flat face and then at the curved face.

Step 1: Refraction at the flat face. The object O is in air ($n_1 = 1$) at a distance $u_1 = 20$ cm from the flat face. The light enters the glass hemisphere ($n_2 = 3/2$). For a plane surface, the radius of curvature is infinite ($R_1 = \infty$). The refraction formula is $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. For a plane surface, this simplifies to $\frac{n_2}{v_1} = \frac{n_1}{u_1}$. Using the sign convention where distances to the right of the surface are positive: $u_1 = +20$ cm. $\frac{3/2}{v_1} = \frac{1}{+20}$ $v_1 = \frac{3}{2} \times 20 = +30$ cm. The first image $I_1$ is formed at a distance of 30 cm to the right of the flat face.

Step 2: Refraction at the curved face. The first image $I_1$ acts as the object for the curved surface. This is the interface between glass ($n_1 = 3/2$) and air ($n_2 = 1$). The observer is viewing from the curved side (air). The object distance $u_2$ is the distance of $I_1$ from the pole of the curved surface. Assuming the flat face is at $x=0$, the curved surface is part of a sphere with its center of curvature at $x=-R = -10$ cm. The pole of the curved surface is at $x=0$. The object $I_1$ is at $x=+30$ cm. Therefore, $u_2 = +30$ cm. The radius of curvature of the curved surface is $R_2 = -10$ cm (negative because the center of curvature is on the left of the pole, which is the side of the incident ray). Applying the refraction formula: $\frac{n_2}{v_2} - \frac{n_1}{u_2} = \frac{n_2 - n_1}{R_2}$ $\frac{1}{v_2} - \frac{3/2}{+30} = \frac{1 - 3/2}{-10}$ $\frac{1}{v_2} - \frac{1}{20} = \frac{-1/2}{-10}$ $\frac{1}{v_2} - \frac{1}{20} = \frac{1}{20}$ $\frac{1}{v_2} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$ $v_2 = +10$ cm.

The final image is formed at a distance of 10 cm to the right (behind) of the flat face.