Question

An illuminated slide is held $\mathbf{44}\mathbf{}\mathbf{cm}$ from a screen. How far from the slide must a lens of focal length$\mathbf{11}\mathbf{cm}$ be placed (between the slide and the screen) to form an image of the slide’s picture on the screen?

Short answer

Distance from the slide at which the lens should be placed is $22\mathrm{cm}$.

Step-by-step solution

Listing the given quantities

Focal length,$\mathrm{f}=11\mathrm{cm}.$

Distance between slide and screen is $44\mathrm{cm}$

Understanding the concepts of lens equation

By using the thin lens equation, we can find the distance at which the lens should be placed.

Formula:

Thin lens equation,

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{p}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{i}}$

Calculations of the distance from the slide at which the lens should be placed

Thin lens equationis

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}$

Since the image is formed on the screen, so the sum of the object distance and the image distance is equal to the distance between the slide and the screen.

As the distance between slide and screen is$44\mathrm{cm}$, i.e.

$\mathrm{p}+\mathrm{i}=44\mathrm{cm}$Or

$\mathrm{i}=44-\mathrm{p}$

Therefore,

$\begin{array}{l}\frac{1}{11}=\frac{1}{\mathrm{p}}+\frac{1}{44-\mathrm{p}}\\ \frac{1}{11}=\frac{(44-\mathrm{p})+\mathrm{p}}{44\mathrm{p}-{\mathrm{p}}^2}\\ \frac{1}{11}=\frac{44}{44\mathrm{p}-{\mathrm{p}}^2}\end{array}$

$44\mathrm{p}-{\mathrm{p}}^2=484$

${\mathrm{p}}^2-44\mathrm{p}+484=0$

After solving the quadratic equation, we get,

$\mathrm{p}=22\mathrm{cm}$

Distance from the slide at which the lens should be placed is $22\mathrm{cm}$