Question

A biconvex lens of focal length 0.2 m acts as divergent lens of power 1d when immersed in liquid. Find refractive index of liquid

Answer 1

Answer:

The refractive index of liquid is 1.01

Explanation:

According to the problem the focal length of the biconvex lens is f= 0.2 m

after immersing in liquid,

the focal length will be, f'= 1/1d = 1m

Now for biconvex lens

1/f = (n-1)(1/r1+ 1/r2)

here n is the refractive index of the biconvex lens =1.5

let n' be the refractive index of the liquid

now for refractive index before immersing and after immersing

f'/f = n-1/n'-1

1/0.2= 1.5-1/n'-1

=>n'-1 = .01

=> n' = 1.01

Answer 2

Answer:

1.667

Explanation:

Power of lens in liquid = 1/f = 1/1 = 1 But the lens is actually a diverging one when put in water hence focus = -1

Focus when lens is normally in air = 0.2 No change as convex is convergent.

Now applying lens makers formula we can get the ratio

Fw/Fa=Ua-1/Uw-1

-1/0.2 = 1.5-1/Uw-1

Uw = 0.9

But we also know Uw=Ul(Refractive index of glass in lens)/ Us(Refractive index of glass in surrounding(here liquid))

0.9 = 1.5 / Us

Us = 1.5/0.9 = 1.67

This is refractive index of liquid