Question
Derive the lens formula,  $\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$ for a concave lens, using the necessary ray diagram.
Two lenses of powers 10 D and – 5 D are placed in contact.
  1. Calculate the power of the new lens.
  2. Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

Answer

Diagram:

Derivation: $\frac{1}{\text{v}} -\frac{1}{\text{u}} = \frac{1}{\text{f}}$

  1. $\text{Power} = \text{P}_{1} + \text{P}_{2}$

$ = 10 - 5 = 5\text{D}$

  1. f = 20 cm

 $\text{m} = \frac{\text{v}}{\text{u}} = 2$

$\frac{1}{\text{f}} =\frac{1}{\text{v}} - \frac{1}{\text{u}}$

$\text{u} = -10\text{cm}.$

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