Question

A biconvex lens with focal length f in air and refractive index of 1.5 is floating on the surface of a deep pond of water (refractive index 1.33). If an object is placed at a height of 2 f vertically above the lens, then the distance between the lens and the image is ?????????

Answer 1

$\frac{1}{f} =(\frac{u'}{u}-1 )(\frac{1}{R1} -\frac{1}{R2} )\\\\\frac{1}{f} =(\frac{1.5}{1}-1 )(\frac{1}{R} -\frac{1}{(-R)})\\\\\frac{2}{f} =\frac{2}{R} \\\\f=R$

Now, when the lens is kept in water, you can consider the lens' surrounding medium to be water with μ=1.33

$\frac{1}{f'} =(\frac{1.5}{1.33}-1 )(\frac{2}{R} )\\\\\frac{1}{f'} =(0.125 )(\frac{2}{f} )\\\\\\\frac{1}{f'} =\frac{1}{4f}$

Applying lens formula with sign convention that above the lens is negative

$\frac{1}{f'} =\frac{1}{v}-\frac{1}{u}\\\\\frac{1}{4f} =\frac{1}{v}-\frac{1}{(-2f)}\\\\\frac{1}{v} =\frac{1}{4f}-\frac{1}{2f}\\\\\frac{1}{v} = \frac{-1}{4f}\\\\v=-4f$

i.e image is formed at a distance of 4f units above the lens in air

Hope this answer helped you :)