Derive the lens formula 1/f = 1/v - 1/u for a concave lens, using the necessary ray diagram.

Question

Derive the lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) for a concave lens, using the necessary ray diagram. 

Two lenses of powers 10D and -5D are placed in contact.

(i) Calculate the power of the new lens.

(ii) Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

Answer 1

Consider a concave lens of focal length f and optical centre C. Let an object AB be placed perpendicular to the principal axis at any suitable distance. The image formed by this lens is always virtual, erect and on the same side of the lens as shown in Fig. 

.

Using sign conventions,

Let CB' = -v, CB = -u, CF = -f

So Eq.(3) becomes

\(\frac{-v}{-u} = \frac{-f+ v}{-f}\)

or -vf = -uf + uv

Dividing both sides by uvf, we get

(i) Power of the combination
= P₁ + P₂ = 10D + (-5D) = 5D

(ii) So focal length of the combination

f = \(\frac{100}{P} = \frac{100}{5} \) = 20 cm

Given m = \(\frac{v}{u}\) so v = mu = 2u

since \(\frac{1}{f} -\frac{1}{u}+\frac{1}{24}\)

or u = 10 cm