What is the magnification of the image when a lens of focal length 10 cm and object distance from the lens is 15 cm?

1. 2
2. 1
3. $\dfrac{1}{2}$
4. 4

2

Reason — As we know, the lens formula is —

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\[1em]$

Given,

f = 10 cm

u = -15 cm

v = ?

Substituting the values in the formula, we get,

$\dfrac{1}{10} = \dfrac{1}{v} – \dfrac{1}{-15} \\[1em] \dfrac{1}{10} = \dfrac{1}{v} + \dfrac{1}{15} \\[1em] \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15} \\[1em] \dfrac{1}{v} = \dfrac{3-2}{30} \\[1em] \dfrac{1}{v} = \dfrac{1}{30} \\[1em] \Rightarrow v = 30 \text{ cm} \\[1em]$

∴ Image distance is 30 cm.

As we know, the formula for magnification of a lens is —

$m = \dfrac{v}{u}$

Given,

v = 30 cm

u = -15 cm

Substituting the values in the formula, we get,

$m = \dfrac{30}{- 15} \\[1em] m = -2$

∴ The magnification is 2, negative sign implies image is inverted.