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A $4 cm$ tall object is placed perpendicular to the principal axis of a convex lens with a focal length of $24 cm$ . The distance of the object from the lens is $16 cm$ . What is the size of the image?

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3 cm

6 cm

12 cm

16 cm

answer is C.

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Detailed Solution

The size of the image is

12 cm

.
Given,
Object distance,

u = - 16 cm

,
Object height

h = 4 cm

and
Focal length,

f = 24 cm

.
Form the lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

Where

f

v

, and

u

are the focal length, image distance, and object distance, respectively.
On arranging the terms, we get

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

On substituting the given values, we get the image distance as

\Rightarrow \frac{1}{v} = \frac{1}{24} + \frac{1}{(- 16)}

\Rightarrow v = - 48 cm

Hence, the distance of the image formed is

- 48 cm

.
Now, it is known that the magnification produced is the ratio of the image distance

(v)

to the object distance

(u)

. It can also be defined as the ratio of the image height

(h')

to the object height

(h)

. Mathematically,

m = \frac{v}{u} = \frac{h'}{h}

On substituting the given values, we get the image height as

(h')

\Rightarrow \frac{v}{u} = \frac{h'}{h}

\Rightarrow h' = \frac{vh}{u}

\Rightarrow h' = \frac{- 48 \times 4}{- 16}

\Rightarrow h' = 12 cm

Hence, the size of the image is

12 cm

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