Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of 15 cm from the pole of the surface. The distance of image from the pole of the surface is ____________ cm.

Solution

Physics sir · 4 steps

Use the single-surface refraction formula, and let the signs of u and R do the work.

1. Set the signs. Light goes from the rarer medium into the denser one, so n₁ = 1.0 and n₂ = 1.5. The object is on the incoming side, so u = -15 cm. The centre of curvature is on the outgoing (denser) side, so R = +30 cm.

   n₂/v - n₁/u = (n₂ - n₁)/R

2. Put in numbers. 1.5/v - 1/-15 = 0.5/30, so 1.5/v = 1/60 - 4/60 = -1/20.
3. Solve for v. v = -1.5 × 20 = -30 cm. The minus sign means the image is virtual and lies in the rarer medium, on the same side as the object.
4. So. The image is 30 cm from the pole.

The common slip: 18. Taking u = +15 cm (or R = -30 cm) flips one term and gives 18 cm, not 30.

The idea to keep. For refraction at one curved surface, fix the direction of light first. Then u is negative for a real object, and R is positive when the centre lies on the side the light goes into.

Answer from the NTA final answer key. Solution by PrepWiser.