Question

A diverging lens of refractive index 1.5 and of focal length 20cm in air has the same radii of curvature for both sides. If it is immersed in a liquid of refractive index 1.7. Calculate the focal length of the lens in the liquid.

Answer 1

Definition. A lens placed in the path of a beam of parallel rays can be called a diverging lens when it causes the rays to diverge after refraction. It is thinner at its center than its edges and always produces a virtual image. A lens with one of its sides converging and the other diverging is known as a meniscus lens ...

Answer 2

Answer:

Explanation:

Refractive index of the lens material, aμg =1.5

Focal length of the lens in air, fair = 15 cm

Using lens makers formula-

1/fair = (aμg-1) [1/R1 -1/R2]

1/15 = (1.5-1)[1/R1 -1/R2]

[1/R1 -1/R2] = 1/7.5

Refractive index of liquid, aμL =1.7

Refractive index of lens glass with respect to liquid, Lμg=aμg /aμL

Lμg= 1.5/1.7 = 0.88

Let the focal length of the lens in liquid is-

1/fliquid = (Lμg-1)[1/R1 -1/R2]

1/fliquid = (0.88-1)[1/7.5]

= -0.015

So,fliquid = -66.67 cm.