Question

A lens is made of flint glass (refractive index $=1.5$ ). When the lens is immersed in a liquid of refractive index $1.25$, the focal length :

• A. increases by a factor of 1.25
• B. increases by a factor of 2.5
• C. increases by a factor of 1.2
• D. decreases by a factor of 1.2

Answer 1

Answer: (B)

increases by a factor of 2.5

Answer 2

Lens-maker's formula is given by $\frac{1}{f}=\left(_{a} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\,\,\, ...(i)$ where ${ }_{ a } \mu_{ g }$ is refractive index of glass w.r.t. air, $R_{1}$ and $R_{2}$ are radii of curvature of two surfaces of lens and $f$ is focal length of the lens. If the lens is immersed in a liquid of refractive index $\mu_{1}$ then $\frac{1}{f_{1}}=\left(_{1} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\,\,\, ...(ii)$ Here, ${ }_{1} \mu_{g}$ is refractive index of glass w.r.t. liquid. Dividing E (i) by E (ii), we have $\frac{f_{1}}{f}=\frac{\left({ }_{a} \mu_{g}-1\right)}{\left({ }_{1} \mu_{g}-1\right)}$ $\Rightarrow \frac{ f _{1}}{ f }=\left(\frac{1.5-1}{\frac{1.5}{1.25}-1}\right)$ $\Rightarrow \frac{ f _{1}}{ f }=\frac{0.5 \times 1.25}{0.25}=2.5$ Hence, focal length increases by a factor of $2.5$.