The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at 2ʳᵈ/3 of the distance of the object from the surface. The wavelength of light inside the surface is 2/3 times the wavelength in air. The radius of the curved surface is x/13 m. The value of 'x' is ___________.

Solution

Physics sir · 4 steps

Get the object distance and refractive index first, then the single-surface formula gives R.

1. Refractive index. The frequency stays the same across the surface, so the wavelength shrinks by the factor μ. Since λ_(glass) = 2/3λₐᵢᵣ, we get μ₂ = 3/2, with μ₁ = 1 for air.
2. Object and image. The real image is at v = +10 m. It is 2/3 of the object distance, so the object is 15 m in front of the surface, giving u = -15 m.

   μ₂/v - μ₁/u = (μ₂ - μ₁)/R

3. Put in numbers. (3/2)/10 - 1/-15 = 3/20 + 1/15 = 13/60. This equals (1/2)/R, so R = 1/2 × 60/13 = 30/13 m.
4. So. Matching R = x/13 m gives x = 30.

The common slip: 78. Taking u = +15 m turns the plus into a minus, giving 1/12 on the left, R = 6 m and x = 78.

The idea to keep. For refraction at one curved surface, use μ₂/v - μ₁/u = (μ₂ - μ₁)/R with signs measured from the surface. A real object always has a negative u.

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