Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :

Solution

Physics sir · 4 steps

The centre C is on the object's side, so R is negative. Get that sign right and the rest follows.

1. Set up signs. Light goes from region I to region II, so measure every distance from the surface and take the direction of light as positive. The object is on the left, so u = -40 cm. The centre C is also on the left, so R = -25 cm.

   μ₂/v - μ₁/u = (μ₂ - μ₁)/R

2. Put in numbers. With μ₁ = 1.25 and μ₂ = 1.4, the formula becomes 1.4/v + 1.25/40 = 0.15/-25. That gives 1.4/v = -0.006 - 0.03125 = -0.03725 cm⁻¹.
3. Solve for v. v = -1.4/0.03725 ≈ -37.58 cm. The minus sign means the image is virtual and lies in region I, on the same side as the object.
4. So. The image is 37.58 cm from the surface, which is option (d).

The idea to keep. For refraction at a curved surface, look at where the centre of curvature is. If it is on the side the light comes from, R is negative, whatever the surface looks like.

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