Question

The focal length of an equiconvex lens is 1.00m when it is placed under water . calculate the focal length of the same lens in air.Take the refractive indices if water and glass as4/3 and3/2 respectively.​

Answer 1

Answer:

recall the lens maker's formula

1/f = (u - 1) ( 1/R1 + 1/R2)

If we use equiconvex lens then R1 = R2 = R

So 1/f = (u-1) * 2/R

Here u is for mu which stands for refractive index of the material of the lens with respect to air medium. And it is given u = 3/2

Plugging we get f = R and so Radius of curvature is 1 m

Now as we place the lens immersed within water the refractive index of glass with respect to the surrounding medium will be (3/2)/(4/3)

So relative refractive index u1 = 9/8

Plugging 1/f1 = (9/8 -1) * 2/1 = 2/8 = 1/4

Hence new focal length would become 4 m

Answer 2

Answer:

Required focal length is 4 m.

Explanation:

By lens maker's formula

$\frac{1}{f} = (u - 1)( \frac{1}{R_1} + \frac{1}{ R_2} )$

For Equiconvex lens then $R_1=R_2=R$

So,

$\frac{1}{f} = (u - 1) \times \frac{2}{R}$

Here u is refractive index.It is given u=$\frac{3}{2}$

Plugging we get $f =R$

and so Radius of curvature is 1 m.

$\frac{1}{f} = (u - 1) \frac{2}{R}$

$\frac{1}{f} = ( \frac{3}{2} - 1) \frac{2}{R} \\ f =R$

Now in water refractive index $= \frac{ \frac{3}{2} }{ \frac{4}{3} } = \frac{9}{8}$

$\frac{1}{f1} = \frac{1}{4}$

So,$f1 = 4 \: \: m$

Required focal length is 4 m.