Q. A concave lens of glass, refractive index 1.5, has both surfaces of same radius of curvature $R$. On immersion in a medium of refractive index 1.75, it will behave as a
A) Convergent lens of focal length 3.5 $R$
B) Convergent lens of focal length 3.0 $R$
C) Divergent lens of focal length 3.5 $R$
D) Divergent lens of focal length 3.0 $R$
Explanation
(a) $\frac{f_{l}}{f_{a}} = \frac{\ _{a}\mu_{g} - 1}{\ _{l}\mu_{g} - 1} = \frac{1.5 - 1}{\frac{1.5}{1.75} - 1} = - \frac{1.75 \times 0.50}{0.25} = - 3.5$
$\therefore f_{l} = - 3.5\ f_{a} \Rightarrow f_{l} = + 3.5R\ \lbrack\because f_{a} = R\rbrack$
Hence on immersing the lens in the liquid, it behaves as a converging lens of focal length $3.5\ R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.