Question

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

Answer 1

Step 1. Given data

Focal length $\mathrm{f}=20\mathrm{cm}$

Distance $=40\mathrm{cm}$

We have to find the maximum value of distance $\mathrm{h}$.

Step 2. Formula used.

A lens is a piece of a refracting medium bounded by two surfaces, at least one of which is a curved surface.

We will calculate the image distance by using the lens formula.

This formula represents the relation between object distance $\mathrm{u}$, image distance $\mathrm{v}$, and focal length $\mathrm{f}$.

$\Rightarrow$ $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$

Here,

$\mathrm{f}=20\mathrm{cm}$

According to the sign of convention, object distance $\mathrm{u}=-40\mathrm{cm}$

$\mathrm{v}=?$

Step 3. Calculate the image distance.

By using the lens formula, we get,

$\frac{1}{20}=\frac{1}{\mathrm{v}}-\frac{1}{-40}$

$\frac{1}{20}=\frac{1}{\mathrm{v}}+\frac{1}{40}$

$\frac{1}{\mathrm{v}}=\frac{1}{20}-\frac{1}{40}$

$\frac{1}{\mathrm{v}}=\frac{1}{40}$

$\mathrm{v}=40\mathrm{cm}$

So, the image distance is $40\mathrm{cm}$, which means the image is formed at the center of curvature.

Draw the ray diagram

Step 4. Find the value of distance.

According to the above figure, $∆\mathrm{CED}$ and $∆\mathrm{CAB}$ are in symmetric form.

So,

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{DC}}$

$\frac{5}{\mathrm{h}}=\frac{40}{20}$

$\frac{5}{\mathrm{h}}=2$

$\mathrm{h}=\frac{5}{2}$

$\mathrm{h}=2.5\mathrm{cm}$

The maximum value of $\mathrm{h}$ to see the image is $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}$.