Question

A bi-convex lens of focal length f₁ is placed in air as shown in Figure 3(a) below. The radii of curvature of its first and second surfaces are R and 2R respectively. The lens is cut along the plane CD. Compare the focal length f₂ of the resulting lens shown in Figure 3(b) with that of the original lens shown in Figure 3(a).

Answer 1

To compare the focal lengths of the two lenses, we apply the Lens Maker’s Formula:

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

Focal length f₁ of the original lens (Figure 3a):

For the bi-convex lens, using the standard sign convention (light traveling from left to right):

R₁ = +R (convex surface)

R₂ = −2R (convex surface)

`1/(f_1) = (mu - 1)(1/R - (-1/(2 R)))`

= `(mu - 1)(1/R + 1/(2 R))` 

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

= `(3(mu - 1))/(2 R)`   ...(i)

Focal length f₂ of the resulting lens (Figure 3b):

When the lens is cut along the plane CD, the resulting lens shown in Figure 3(b) is plano-convex.

The first surface is now the plane CD, so R₁ = ∞

The second surface remains the original curved surface, so R₂ = −2 R

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

= `1/(f_2) = (mu - 1)(0 + 1/(2 R))`

= `(mu - 1)/(2 R)`    ...(ii)

By dividing Equation (i) by Equation (ii):

`((1/(f_1)))/((1/(f_2))) = ((3(mu - 1))/(2 R))/((mu - 1)/(2 R))`

`(f_2)/(f_1) = (3(mu - 1))/(2 R) xx (2 R)/(mu - 1)`

`(f_2)/(f_1) = 3`

f₂ = 3 f₁

∴ The focal length of the resulting lens (f₂) is three times the focal length of the original lens (f₁).