Question

A thin plano-convex lens and a thin equi-concave lens are kept coaxially in contact as shown in the figure. Assuming both the lenses are made of glass of refractive index μ, and R is the radius of curvature of each curved surface, the focal length of the combination is:

Options

• `R/(mu - 1)`

• `-R/(mu - 1)`

• `(2R)/(mu - 1)`

• `-(2R)/(mu - 1)`

Answer 1

`bb(-R/(mu - 1))`

Explanation:

For thin lenses placed in contact:

`1/f_(eq) = 1/f_1 + 1/f_2`

Use lens maker’s formula:

`1/f_1 = (mu - 1)(1/R_1 - 1/R_2)`

It has one curved surface and one plane surface.

`1/f_1 = (mu - 1)(1/R - 0)`

`f_1 = R/(mu - 1)`

Equi-concave lens, both surfaces concave:

R₁ = −R

R₂ = +R

`1/f_2 = (mu - 1)((-1)/R - 1/R)`

`1/f_2 = -(2(mu - 1))/R`

`f_2 = -R/(2(mu - 1))` 

Equivalent focal length

`1/f_(eq) = (mu - 1)/R - (2(mu - 1))/R`

`1/f_(eq) = -(mu - 1)/R`

∴ `f_(eq) = -R/(mu - 1)`