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Physics Question on Ray optics and optical instruments

A convex lens of focal length 20 cm made of glass of refractive index 1.5 is immersed in water having refractive index 1.33. The change in the focal length of lens is

A

62.2 cm

B

5.82 cm

C

58.2 cm

D

6.22 cm

Answer

58.2 cm

Explanation

Solution

When the lens is in air
1fa=(μg−1)(1R1−1R2)\frac{1}{f_{a}} =\left(\mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)
120=(1.5−1)(1R1−1R2)\frac{1}{20} = \left(1.5 -1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) ...(i)
When lens is in water,
1fw=(μgμw−1)(1R1−1R2)\frac{1}{f_{w}} =\left(\frac{\mu_{g}}{\mu_{w}} - 1\right)\left(\frac{1}{R_{1}} -\frac{1}{R_{2}}\right)
or 1fw=(1.5−1.331.33)(1R1−1R2)\frac{1}{f_{w}} =\left(\frac{1.5-1.33}{1.33}\right)\left(\frac{1}{R_{1}} -\frac{1}{R_{2}}\right) ....(ii)
Divide (i) by (ii), we get
or, fw20=(1.5−1)(1.331.5−1.33)\frac{f_{w}}{20} =\left(1.5 -1\right)\left(\frac{1.33}{1.5-1.33}\right)
or, fw=20×0.5×1.330.17=78.2cmf_{w} = 20 \times0.5 \times\frac{1.33}{0.17} ={ 78.2 cm}
The change in focal length = 78.2−20=58.2 cm{78.2 - 20 = 58.2 \, cm}