JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
A convex lens of refractive index \(1.5\) and focal length \(18 cm\) in air is immersed in water. The change in focal length of the lens will be \(...........cm\).\(\left(\right.\) Given refractive index of water \(\left.=\frac{4}{3}\right)\)
- A \(55\)
- B \(53\)
- C \(54\)
- D \(52\)
Answer & Solution
Correct Answer
(C) \(54\)
Step-by-step Solution
Detailed explanation
\(\frac{ I }{ f _{ H _2 O }}=\left(\frac{\mu_{ g }}{\mu_{ H _2 O }}-1\right)\left(\frac{2}{ R }\right)\) \(=\frac{1}{8}\left(\frac{2}{ R }\right)\) \(=\frac{1}{\left(4 f _{\text {air }}\right)}\) So, \(f _{ H _2 O }=4 f _{ air }=72\,cm\) So change in focal length…
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