Question

A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, what will be its new focal length?

How does the nature of the lens change, if this lens is immersed in a liquid of refractive index 1.8?

Answer 1

From (1) and (2), we get

f = 52 cm

When the converging lens is immersed in a liquid of refractive index 1.3, it will behave like divergent lens of focval length say f'

∴ \(\frac{1}{f'} = (\frac{1.6}{1.8} - 1) (\frac{1}{R_1} - \frac{1}{R_2}) \)

or \(\frac{1}{f'} = -\frac{1}{9} (\frac{1}{R_1} - \frac{1}{R_2})\) ...........(3)

From (1) and (3)

f' = -108 cm

i.e. the focal length of divergent lens is -108 cm.