A concave lens has a focal length of 30 cm. Find the position and magnification (m) of the image for an object placed in front of it at a distance of 30 cm. State whether the image is real or virtual ?

As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given,

f = - 30 cm

u = - 30 cm

Substituting the values in the formula, we get,

$\dfrac{1}{v} – \dfrac{1}{-30} = \dfrac{1}{-30} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{30} = - \dfrac{1}{30} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{30} - \dfrac{1}{30} \\[0.5em] \dfrac{1}{v} = \dfrac{-1-1}{30} \\[0.5em] \dfrac{1}{v} = -\dfrac{2}{30} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{15} \\[0.5em] \Rightarrow v = -15$

Therefore, the image in formed 15 cm in front of the lens.

ii) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$ Given,

u = - 30 cm

v = - 15 cm

Substituting the values in the formula, we get,

$m = \dfrac{-15}{-30} \\[0.5em] m = 0.5 \\[0.5em]$

Therefore, the magnification is +0.5

The image is virtual as the magnification is positive.