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

The Correct Option is B

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 $