Question 269: Ray Optics

Question

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is $\_\_\_\_$ cm.

Accepted Answer

For a thin lens to form images of the same size at two different object positions, one image must be real and the other must be virtual. This typically occurs with a convex lens.
If the object distance is $u$, focal length of lens is $f$ then lens formula is given as,
$$ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{u+f}{u f} $$
The magnification is given as $m=\frac{v}{u} \Rightarrow m=\frac{1}{u} 	imes\left[\frac{u f}{u+f}ight]=\frac{f}{f+u}$
Where:
f is the focal length.
u is the object distance (using sign convention, u is negative).
Since the sizes (magnitudes of magnification) are equal but the nature of the images is different:
$$ \left|\mathrm{m}_1ight|=\left|\mathrm{m}_2ight| \Rightarrow \mathrm{m}_1=-\mathrm{m}_2 $$
Given:
1. Position $1=\mathrm{u}_1=-8 \mathrm{~cm}$
2. Position $2=\mathrm{u}_2=-24 \mathrm{~cm}$
$\frac{f}{f+(-8)}=-\left(\frac{f}{f+(-24)}ight)$
$$\Rightarrow $$ $\frac{1}{f-8}=-\frac{1}{f-24}$
$$\Rightarrow $$ $f-24=-(f-8)$
$$\Rightarrow $$ $f-24=-f+8$
$$\Rightarrow $$ $2 \mathrm{f}=32 $
$$\Rightarrow $$ $ \mathrm{f}=16 \mathrm{~cm}$
The focal length of the lens that produces images of equal size at object distances of 8 cm and 24 cm is 16 cm.
Hence, the correct answer is 16.

Ray Optics

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is \_\_\_\_ cm.

Choose the correct answer: