Question

A thin converging lens is made up of glass of refractive index 1.5. It acts like a concave lens of focal length 50 cm, when immersed in a liquid of refractive index $\left( \frac{15}{8} \right)$ . The focal length of the converging lens in air is, in metre:

• A. 0.15
• B. 0.2
• C. 0.25
• D. 0.4

Answer 1

Answer: (B)

0.2

Answer 2

Given: Focal length of concave lens $=50\,cm=0.5\,m$ Refractive index of lens of ${{\mu }_{lens}}=1.5$ Refractive index of liquid ${{\mu }_{liquid}}=\frac{15}{8}$ Now, from lens markers formula $\frac{1}{{{f}_{lens}}=}({{\mu }_{lens}}-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{1}}} \right)$ $\Rightarrow$ $\frac{1}{{{f}_{lens}}}=(1.5-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)$ $\Rightarrow$ $\frac{1}{{{f}_{lens}}\times 0.5}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}$ $\Rightarrow$ $\frac{2}{{{f}_{lens}}}=\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)$ ?(i) Now, the lens is immersed in liquid of refractive index $\frac{15}{8}$ then, $\frac{1}{f}=\left( \frac{{{\mu }_{lens}}}{{{\mu }_{liquid}}}-1 \right)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)$ $\Rightarrow$ $-\frac{1}{0.5}=\left( \frac{15}{15/8}-1 \right)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)$ $\Rightarrow$ $-2=(0.8-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)$ $\Rightarrow$ $10=\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)$ ?(ii) Putting the value of $\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}$ from E(ii) in E(i) , we get $\frac{2}{{{f}_{lens}}}=10\Rightarrow 10{{f}_{lens}}=2\Rightarrow {{f}_{lens}}=0.2\,m$