Question

A point object is placed on the principle axis of a convex lens of focal length f at a distance of $2$f to the left of it. The diameter of the lens is d. An eye is placed at a distance of $3$f to the right of the lens and a distance h below the optic axis. The maximum value of h to see the image is?

• A. d
• B. $d/2$
• C. $d/3$
• D. $d/4$

Answer 1

The correct option is B $d/2$

Given: A point object is placed on the principle axis of a convex lens of focal length $f$ at a distance of $2f$ to the left of it. The diameter of the lens is $d$. An eye is placed at a distance of $3f$ to the right of the lens and a distance $h$ below the optic axis.

To find the maximum value of $h$ to see the image

Solution:

As per the given criteria,

object distance, $u=-2f$

focal length, $f$

Applying the lens formula, we get

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}⟹\frac{1}{v}=\frac{1}{f}+\frac{1}{u}⟹\frac{1}{v}=\frac{1}{f}+\frac{1}{(-2f)}⟹\frac{1}{v}=\frac{2-1}{2f}⟹v=2f$

So the image is formed at the curvature.

so we get a ray diagram as shown in above fig.

In the above fig.

$\Delta CDE$ and $\Delta CAB$ are symmetric.

$BC=EB-BC=3f-2f=f$

So $\frac{AB}{DB}=\frac{BC}{EC}⟹\frac{h}{d}=\frac{f}{2f}⟹h=\frac{d}{2}$

Hence the maximum value of $h$ to see the image is $\frac{d}{2}$