A lens forms the image of an object placed at a distance of 45 cm from it on a screen placed at a distance 90 cm on the other side of it.

(a) Name the kind of lens.

(b) Find : (i) the focal length of lens, and (ii) the magnification of the image.

(a) As the image is formed on the other side of the lens, so the image is real. Hence, the lens is convex.

(b) (i) As we know, the lens formula is —

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

Given,

u = – 45 cm

v = + 90 cm

Substituting the values in the formula, we get,

$\dfrac{1}{90} – \dfrac{1}{-45} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{90} + \dfrac{1}{45} = \dfrac{1}{f} \\[0.5em] \dfrac{1+2}{90} = \dfrac{1}{f} \\[0.5em] \dfrac{3}{90} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{30} = \dfrac{1}{f}\\[0.5em] \Rightarrow f = 30cm \\[0.5em]$

Therefore, focal length of the lens is 30 cm.

ii) As we know,

the formula for magnification of a lens is —

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

u = –45 cm

v = +90 cm

Substituting the values in the formula, we get,

$m = \dfrac{90}{-45} \\[0.5em] m = -2 \\[0.5em]$

Therefore, the magnification is -2.