If a lens of refractive index μ is immersed in a medium of refractive index μ1, then its focal length in medium is given by fm1=(mμl−1(R11)−R21) If fa is the focal length of lens in air, then fa1=(aμl−1(R11−R21)⇒fafm=(mμl−1)(aμl−1) If μ1>μ, then fm and fa have opposite signs and the nature of lens changes i.e. a convex lens diverges the light rays and concave lens converges the light rays. Thus given option (a) is correct.
The refractive index of the material of a concave lens is μ. It is immersed in a medium of refractive index μ_1. A parallel beam of light is incident…
JEE Main 2014 — Physics Optics
2014mcqeasy
The refractive index of the material of a concave lens is μ. It is immersed in a medium of refractive index μ1. A parallel beam of light is incident on the lens. The path of the emergent rays when μ1>μ is:
Official previous-year question
Held on 12 Apr 2014 · Verified 6 Jul 2026.
Options
- A
- B
- C
- D
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