A convex lens of refractive index 1.5 has a focal length of 18 cm in air. Calculate the change in its focal length when it is immersed in water of refractive index 4/3.

Solution

Given, a convex lens.
Refractive index of lens, ${}^{a}μ_{g}$ = 1.5
Focal length of lens in air, f_a = 18
Refractive index of water, ${}^{a}μ_{w} = \frac{4}{3}$

For the lens in air,
$\frac{1}{f_{a}} = ({}_{a}μ_{g} - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

   $\frac{1}{18} = (1.5 - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{R_{1}} - \frac{1}{R_{2}} = \frac{1}{4}$

When the lens is immersed in water
$\frac{1}{f_{w}} = (\frac{{}^{a}μ_{g}}{{}_{a}μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

   $= (\frac{1.5}{4 / 3} - 1) \times \frac{1}{4} = \frac{1}{8} \times \frac{1}{4} = \frac{1}{32}$

Thus, $f_{w} = 32 cm .$

Hence, focal length changes from 18 to 32.

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Wave Optics

A convex lens of refractive index 1.5 has a focal length of 18 cm in air. Calculate the change in its focal length when it is immersed in water of refractive index 4/3.

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