Optics 3 Question 30
30. A thin lens of refractive index 1.5 has a focal length of $15 cm$ in air. When the lens is placed in a medium of refractive index $4 / 3$, its focal length will become …… cm.
(1987, 2M)
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Answer:
Correct Answer: 30. 60
Solution:
$$ \begin{aligned} \frac{1}{f _{\text {air }}} & =(1.5-1) \frac{1}{R _1}-\frac{1}{R _2} \\ \frac{1}{f _{\text {medium }}} & =\frac{1.5}{4 / 3}-1 \frac{1}{R _1}-\frac{1}{R _2} \end{aligned} $$
Dividing Eq. (i) by Eq. (ii), we get
$$ \begin{aligned} \frac{f _{\text {medium }}}{f _{\text {air }}} & =4 \\ \therefore \quad f _{\text {medium }} & =4 f _{\text {air }}=4 \times 15 \\ & =60 cm \end{aligned} $$
