NEET 2019 · PhysicsPrevious Year Question
Two similar thin equi-convex lenses of focal length f each are kept coaxially in contact with each other such that the focal length of the combination is F₁. When the space between the two lenses is filled with glycerin (which has the same refractive index, µ = 1.5, as that of glass) then the equivalent focal length is F₂. The ratio F₁ : F₂ will be
Answer: (B) 1 : 2. Answer: (B) 1 : 2 Solution: For one equi-convex lens with radii of magnitude R: 1/f = (µ − 1)(2/R) = (1.5 − 1)(2/R) = 1/R Case 1 — two lenses in contact (air between): 1/F₁ = 1/f + 1/f = 2/f So F₁ = f/2.
- A.2 : 1
- B.1 : 2✓
- C.2 : 3
- D.3 : 4
Correct Answer
(B) 1 : 2
Solution & Explanation
Answer: (B) 1 : 2 Solution: For one equi-convex lens with radii of magnitude R: 1/f = (µ − 1)(2/R) = (1.5 − 1)(2/R) = 1/R Case 1 — two lenses in contact (air between): 1/F₁ = 1/f + 1/f = 2/f So F₁ = f/2. Case 2 — glycerin (same µ = 1.5 as glass) fills the gap. The glycerin forms an equi-concave lens between the two equi-convex lenses, with radii −R and +R: 1/f_g = (1.5 − 1)(−2/R) = −1/R = −1/f Now three lenses in contact: 1/F₂ = 1/f + (−1/f) + 1/f = 1/f So F₂ = f. Therefore F₁ : F₂ = (f/2) : f = 1 : 2.
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