A point object O is placed on the principal axis of a convex lens of focal length f= 20 cm at a distance of 40 cm to the left of it. The aperture of the lens is 10 cm. An eye is placed 60 cm to the right of the lens and a distance h below the principal axis. The maximum value of h to see the image is 

The lens formula is always used in situations of this kind in optics. Using this formula, we can get the image distance in this case. The value of h is then determined by using the straightforward mathematical symmetry of triangles.

Given: A point object O is positioned 40 cm to the left of a convex lens with a focal length of 20 cm on its primary axis. The lens has a 10 centimetre diameter. An eye is positioned h below the primary axis and 60 cm to the right of the lens.
To see the picture, find the highest value of h.

Considering the stated requirements,
object distance, $\mathrm{u}=-40 \mathrm{cm}$
focal length, $\mathrm{f}=20 \mathrm{cm}$
When the lens formula is used, we obtain
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$
$\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}$
$\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{20}+\frac{1}{(-40)}$
$\Rightarrow \frac{1}{\mathrm{v}}=\frac{2-1}{40}$
$\Rightarrow \mathrm{v}=40 \mathrm{cm}$
As a result, the picture is created at the curvature.
As a result, we obtain the ray diagram in the fig. above. In the figure above,
$∆\mathrm{CDE}$ and $∆\mathrm{CAB}$ are symmetric.

$\text{ So }\frac{\mathrm{AB}}{\mathrm{DB}}=\frac{\mathrm{BC}}{\mathrm{EC}}$
$\Rightarrow \frac{\mathrm{h}}{5}=\frac{20}{40}$
$\Rightarrow \mathrm{h}=\frac{5}{2}=2.5 \mathrm{cm}$
the highest value of h needed to view the picture is $2.5 \mathrm{cm}$.

Hence, the correct answer is option B.