Question

A convex lens of refractive index $\frac{3}{2}$ has a power of $2.5\, D$ in air. If it is placed in a liquid of refractive index $2$, then the new power of the lens is

• A. - 1.25 D
• B. - 1.5 D
• C. 1.25 D
• D. 1.5 D

Answer 1

Answer: (A)

- 1.25 D

Answer 2

Focal length of a convex lens having power $2.5 D,=\frac{1}{2.5} m$ Also focal length of a lens in a medium of refractive index $\mu$ is given by $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ $\Rightarrow 2.5=\frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \quad \ldots(i) \,\,\,\,\,$ (in air) $\Rightarrow \frac{1}{f'}=\left(\frac{3}{4}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ldots(i i) \,\,\,\,\left[\because l \mu_{g}=\frac{3}{4}\right]$ (in liquid) Dividing the two, $2.5 f '=\frac{0.5}{-0.25}$ $\Rightarrow \frac{1}{f'}=\frac{-5}{25 \times 0.25}=-1.25 D$