An equiconvex lens has focal length +10 cm in air; glass refractive index is 1.5. Liquid refractive index increases with time as μ(t) = 1 + t/10 (t in seconds). The immersed lens has focal length −20 cm at t = 5n s. Find n.
Physics · Mixed Chapter
JEE Main Practice

Question

An equiconvex lens has focal length +10 cm in air; glass refractive index is 1.5. Liquid refractive index increases with time as μ(t) = 1 + t/10 (t in seconds). The immersed lens has focal length −20 cm at t = 5n s. Find n.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Lens-maker relation in medium is 1/f = (μg/μₗ − 1)(1/R₁ − 1/R₂). For an equiconvex lens, 1/R₁ − 1/R₂ = 2/R. Step 2 — In air, 1/10 = (1.5 − 1) 2/R ⇒ R = 10 cm. Step 3 — For f = −20 cm, −1/20 = (1.5/μₗ − 1)(2/10) ⇒ 1.5/μₗ = 0.75 ⇒ μₗ = 2. Step 4 — Liquid index is μₗ = 1 + t/10, so t = 10 s. Since t = 5n, n = 2.