JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
A convex lens of refractive index 1.5 and focal length \(f=18 cm\) is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is \(\alpha \times f\). The value of \(\alpha\) is ___________.
(refractive index of water \(=4 / 3\) )
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
\(\frac{1}{ f _{\text {Air }}}=\left(\frac{1.5-1}{1}\right)\left(\frac{1}{ R _1}+\frac{1}{ R _2}\right)\) \(\frac{1}{ f _{\text {water }}}=\left(\frac{1.5-4 / 3}{4 / 3}\right)\left(\frac{1}{ R _1}+\frac{1}{ R _2}\right)\)…
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