Physics Circle home Student Login

Home›Chapters›Ray Optics PYQs›Spherical Surfaces & Thin Lenses›Question 12

MHT CET 2026 (18 Apr, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from MHT CET 2026 (18 Apr, Shift 2)

Q12MHT CET 2026 · 18 APR · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

Refractive index of a glass convex lens is 1.51.5. The radius of curvature of each of the two surfaces of the lens is 40 cm. The ratio of the power of the lens when immersed in a liquid of refractive index 1.251.25 to that when placed in air is

Show answer and solution

Answer: (b)

  1. PlPa=μμl−1μ−1\frac{P_l}{P_a} = \frac{\frac{\mu}{\mu_l} - 1}{\mu - 1} (the radii cancel).
  2. 1.51.25−1=0.2\frac{1.5}{1.25} - 1 = 0.2, so PlPa=0.20.5=25\frac{P_l}{P_a} = \frac{0.2}{0.5} = \frac{2}{5}
  3. So the answer is (b) 2:52 : 5.
Report a mistake

More questions on Spherical Surfaces & Thin Lenses