Refraction Through a Lens — Question 7

(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.