An object placed in front of a convex lens, forms an image of same size on a screen. Moving the object 12 cm closer to the lens results in the formation of a real image which is three times the size of the object. Calculate the focal length of the lens.

(a) Initially,

Let, initial object distance be u, image distance be v and focal length be f.

According to question,

Since the image is formed on a screen so it is real and inverted and it is also of the same size as the object which means magnification produced by the lens is -1.

Magnification (m) of a lens is given by,

$\text m = \dfrac{\text {v}}{\text {u}} \\[1em] \Rightarrow -1 = \dfrac{\text {v}}{\text {u}} \\[1em] \Rightarrow \text {u} = -\text {v}$

From lens formula,

$\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \\[1em] \dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{(-\text v)} \\[1em] \dfrac{1}{\text f} = \dfrac{1}{\text v} + \dfrac{1}{\text v} \\[1em] \dfrac{1}{\text f} = \dfrac{2}{\text v} \\[1em] \Rightarrow \text f = \dfrac{\text v}{2}$

Let, final object distance be u' and final image distance be v'.

According to question,

u' = u + 12 and m' = -3

Magnification (m') of a lens is given by,

$\text m' = \dfrac{\text {v'}}{\text {u'}} \\[1em] \Rightarrow -3 = \dfrac{\text {v'}}{\text {u'}} \\[1em] \Rightarrow \text {u'} = -\dfrac{\text {v'}}{3} \\[1em] \Rightarrow \text {u} + 12 = -\dfrac{\text {v'}}{3} \\[1em] \Rightarrow -\text {v} + 12 = -\dfrac{\text {v'}}{3} \\[1em] \Rightarrow -3(\text {v} - 12) = -\text {v'} \\[1em] \Rightarrow 3(\text {v} - 12) = \text {v'} \\[1em] \Rightarrow 3\text {v} - 36 = \text {v'} \\[1em]$

From lens formula,

$\dfrac{1}{\text f} = \dfrac{1}{\text v'} - \dfrac{1}{\text u'} \\[1em] \dfrac{1}{\text f} = \dfrac{1}{\text v'} - \dfrac{3}{-\text v'} \\[1em] \dfrac{1}{\text f} = \dfrac{1}{\text v'} + \dfrac{3}{\text v'} \\[1em] \dfrac{1}{\text f} = \dfrac{4}{\text v'} \\[1em] \Rightarrow \text f = \dfrac{\text v'}{4} \\[1em] \Rightarrow \text f = \dfrac{3\text {v} - 36}{4} \\[1em] \Rightarrow \text f = \dfrac{3\text {v}}{4} - \dfrac{36}{4} \\[1em] \Rightarrow \text f = \dfrac{3\text {f}}{2} - 9 \\[1em] \Rightarrow \dfrac{3\text {f}}{2} - \text f = 9 \\[1em] \Rightarrow \dfrac{3\text {f}}{2} - \dfrac {2\text f}{2} = 9 \\[1em] \Rightarrow \dfrac {\text f}{2} = 9 \\[1em] \Rightarrow \text f = 2\times 9 \\[1em] \Rightarrow \text f = 18 \text { cm}$

Hence, the focal length of the lens is 18 cm.