A lens of focal length f = 18 cm has a refractive index 3/2. It is immersed in water of refractive index 4/3. The change in focal length is λf. Find the value of λ.

Solution:

Using the lens maker formula:

1/f = (μ_lens/μ_medium − 1) (1/R₁ − 1/R₂)

For air:

1/18 = (3/2 − 1)K

1/18 = (1/2)K

K = 1/9

For water:

1/f' = [(3/2)/(4/3) − 1](1/9)

= (9/8 − 1)(1/9)

= (1/8)(1/9)

= 1/72

f' = 72 cm

Change in focal length:

72 − 18 = 54 cm

54 = λ × 18

λ = 3

Answer: 3