A spherical surface of radius of curvature R, separates air from glass (refractive index =1.5 ). The centre of curvature is in the glass medium. A point object ' O ' placed in air on the optic axis of the surface, so that its real image is formed at 'I' inside glass. The line OI intersects the spherical surface at P and P O=P I. The distance P O equals to

Solution

Physics sir · 3 steps

Use the single-surface refraction formula with u = -x and v = +x, then solve for x.

1. Set up signs. Measure everything from the pole P, taking light to travel from air into glass. Let PO = PI = x. The object is in air, so u = -x. The real image is inside glass, so v = +x. The centre of curvature is in the glass, so R is positive.

   μ₂/v - μ₁/u = (μ₂ - μ₁)/R

2. Put in numbers. With μ₁ = 1 and μ₂ = 1.5: 1.5/x - 1/-x = 0.5/R, so 1.5/x + 1/x = 2.5/x = 0.5/R.
3. So. x = 2.5R/0.5 = 5R, so PO = 5R, which is option (a).

The idea to keep. For refraction at one curved surface, fix the signs first: object on the incoming side is negative, real image beyond the surface is positive, and R is positive when the centre lies on the outgoing side.

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