For a certain lens, the magnification of an object when placed at a distance of 0.15 m is twice of the magnification produced, when the distance was 0.2 m . If in both the situations a real image is f

Question

For a certain lens, the magnification of an object when placed at a distance of 0.15 m is twice of the magnification produced, when the distance was 0.2 m . If in both the situations a real image is formed, then what is the focal length (in cm ) of the lens?

TestHub · Accepted Answer

Let as shown, 1 and 2 are positions of objects and images in two different situations.

Object

It is given

$|\frac{v ₁}{u ₁}| = 2 |\frac{v ₂}{u ₂}|$

Here, $u ₁ = - 15 cm, u ₂ = - 20 cm$

$∴ v ₁ = 2 v ₂ \times \frac{u ₁}{u ₂} = 2 v ₂ \times \frac{15}{20} = \frac{3}{2} v ₂$

$n o w, \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$∴ \frac{1}{f} = \frac{1}{v ₁} - \frac{1}{u ₁} a n d \frac{1}{f} = \frac{1}{v ₂} - \frac{1}{u ₂}$

$s o, \frac{1}{v_{1}} - \frac{1}{u_{1}} = \frac{1}{v_{2}} - \frac{1}{u_{2}}$

$⟹ \frac{2}{2 v ₂} + \frac{1}{15} = \frac{1}{v ₂} + \frac{1}{20}$

$⟹ v = 20 cm$

$∴ \frac{v ₁}{u ₁} = 2 \frac{v ₂}{u ₂} = 2 \times \frac{20}{20} = 2$

$⟹ v ₁ = 2 u ₁ = 2 \times 15 = 30 cm$

Therefore, $\frac{1}{f} = \frac{1}{v ₁} - \frac{1}{u_{1}} = \frac{1}{15} + \frac{1}{30} = \frac{3}{30}$

$∴ f = 10 cm$

Physics - Geometrical Optics Question with Solution | TestHub

Doubts & Discussion