JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
The radius of curvature of each surface of a convex lens having refractive index \(1.8\) is \(20\,cm\). The lens is now immersed in a liquid of refractive index \(1.5\). The ratio of power of lens in air to its power in the liquid will be \(x : 1\). The value of \(x\) is \(.....\)
- A \(3\)
- B \(4\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(P=(1.8-1)\left(\frac{1}{20}+\frac{1}{20}\right)\) by lens maker's formula \(P^{\prime}=\left(\frac{1.8}{1.5}-1\right)\left(\frac{1}{20}+\frac{1}{20}\right)\) Dividing \(\frac{ P }{ P ^{\prime}}=\frac{0.8}{1.2-1}=4\)
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