Question

There is a glass sphere of refractive index \(1.5\), on which a parallel beam of light falls. Find the distance of the final converging point of the emergent rays from the centre of the sphere. Radius of the sphere is \(50\,\text{cm}\).

• A. \(75\,\text{cm}\)
• B. \(70\,\text{cm}\)
• C. \(80\,\text{cm}\)
• D. \(65\,\text{cm}\)

Hint:

For optical problems involving spheres:
Always apply refraction formula separately for each surface
Carefully determine the sign of radius and object distance
Final distance is often asked from the centre, not from the surface

Answer 1

The Correct Option is A

Concept:
When a parallel beam of light falls on a transparent spherical object, refraction occurs at both spherical surfaces. The final image is obtained after applying refraction successively at:
The first spherical surface (air to glass)
The second spherical surface (glass to air) For refraction at a spherical surface, the formula used is: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \] where \(n_1, n_2\) = refractive indices, \(u\) = object distance, \(v\) = image distance, \(R\) = radius of curvature.

Step 1: Refraction at the first surface. Parallel rays imply: \[ u_1 = \infty \] For the first surface: \[ n_1 = 1,\quad n_2 = 1.5,\quad R_1 = +50\,\text{cm} \] Applying the formula: \[ \frac{1.5}{v_1} - \frac{1}{\infty} = \frac{1.5 - 1}{50} \] \[ \frac{1.5}{v_1} = \frac{0.5}{50} \] \[ v_1 = \frac{1.5 \times 50}{0.5} = 150\,\text{cm} \] Thus, the image formed by the first surface is \(150\,\text{cm}\) inside the glass from the first surface.
Step 2: Locate the object for the second surface. The thickness of the sphere (diameter): \[ 2R = 100\,\text{cm} \] Distance of the image from the second surface: \[ u_2 = 150 - 100 = 50\,\text{cm} \] Since the image lies to the right of the second surface, it acts as a virtual object: \[ u_2 = +50\,\text{cm} \]
Step 3: Refraction at the second surface. For the second surface: \[ n_1 = 1.5,\quad n_2 = 1,\quad R_2 = -50\,\text{cm} \] Apply refraction formula: \[ \frac{1}{v_2} - \frac{1.5}{50} = \frac{1 - 1.5}{-50} \] \[ \frac{1}{v_2} - 0.03 = \frac{-0.5}{-50} \] \[ \frac{1}{v_2} - 0.03 = 0.01 \] \[ \frac{1}{v_2} = 0.04 \Rightarrow v_2 = 25\,\text{cm} \] This distance is measured from the second surface.
Step 4: Find distance from the centre of the sphere. Distance of centre from second surface: \[ R = 50\,\text{cm} \] Hence, distance of final image from centre: \[ 50 + 25 = 75\,\text{cm} \] \[ \boxed{\text{Final converging point is } 75\,\text{cm from the centre of the sphere}} \]