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Q.
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
JIPMERJIPMER 2012Ray Optics and Optical Instruments
Solution:
When the lens is in air
$\frac{1}{f_{a}} =\left(\mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$ \frac{1}{20} = \left(1.5 -1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ ...(i)
When lens is in water,
$\frac{1}{f_{w}} =\left(\frac{\mu_{g}}{\mu_{w}} - 1\right)\left(\frac{1}{R_{1}} -\frac{1}{R_{2}}\right)$
or $\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, $\frac{f_{w}}{20} =\left(1.5 -1\right)\left(\frac{1.33}{1.5-1.33}\right) $
or, $f_{w} = 20 \times0.5 \times\frac{1.33}{0.17} = 78.2 cm$
The change in focal length = $78.2 - 20 = 58.2 \, cm$