Question

A student focused the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as:

Position of candle = $12.0\mathrm{cm}$.

Position of convex lens = $50.0\mathrm{cm}$.

Position of the screen = $88.0\mathrm{cm}$.

What will be the nature of the image formed if he further shifts the candle towards the lens?

Note: Candle has been already shifted to the focus of the lens.

Answer 1

Step 1: Given data

Position of candle = $12.0\mathrm{cm}$.

Position of convex lens = $50.0\mathrm{cm}$.

Position of the screen = $88.0\mathrm{cm}$.

Step 2: Formula used

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

Here,

$\mathrm{f}$ is the focal length.

$\mathrm{u}$ is the object's distance.

$\mathrm{v}$ is the distance of the image formed on the screen to the lens.

Step 3: Calculation and explanation

From the above diagram,

The object's distance from the lens, $\mathrm{u}=50-12=-38\mathrm{cm}$

The distance of the image formed on the screen to the lens, $\mathrm{v}=88-50=38\mathrm{cm}$

By using the lens formula, the focal length will be:

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

By substituting the values, we get

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{38}-\frac{1}{-38} \\ \frac{1}{\mathrm{f}}=\frac{1}{38}+\frac{1}{38} \\ \frac{1}{\mathrm{f}}=\frac{2}{38} \\ \frac{1}{\mathrm{f}}=\frac{1}{19} \\ \mathrm{f}=19\mathrm{cm} \end{gathered}$$

1. The candle has been already shifted to the focus of the lens and if it is further shifted, the candle comes in between focus and the optical center of the lens.
2. In this case, the image formed is virtual, enlarged, and erect.
3. It is formed on the same side of the lens where the candle is placed.

An illustrative image of the given case is shown below.