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A $4 cm$ tall object is placed perpendicular to the principal axis of a convex lens of focal length $24 cm$ . the distance of the object from the lens is $16 cm$ . Then the size of the image will be__

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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 will be

12 cm

.
Given,
The height of the object

h_{o} = + 4 cm

The focal length of the lens,

f = + 24 cm

[Positive: as the lens is convex]
Distance of the object

u = - 16 cm

Leth the distance of the image is

= v

By using the lens formula,

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

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

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

\Rightarrow \frac{1}{v} = \frac{1}{24} - \frac{1}{16}

\Rightarrow \frac{1}{v} = \frac{2 - 3}{48}

\Rightarrow \frac{1}{v} = \frac{- 1}{48}

\Rightarrow v = - 48

The negative size indicates that the image will be formed on the same side as that of the object.
Let, the size of the image be H, then by the formula of magnification(

m

) for the lens,

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

\Rightarrow \frac{H}{4} = \frac{- 48}{- 16}

\Rightarrow H = 4 \times \frac{48}{16}

\Rightarrow H = 12 cm

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