Question

A convex lens forms a real and inverted image of a needle at a distance of $25cm$ from the lens. If the image is of the same size as that of the needle, then where should the needle be placed in front of the lens? Also calculate the power of the lens.

Answer 1

Step 1 - Given data.

The image distance $\left(v\right)$ from the lens is equal to $25cm$.

Step 2 - Formulae Used.
The power of the lens is given as: $P=\frac{1}{f}$. Where $P$ is the power of the lens and $f$ is the focal length of the lens in metres.
To find the power of the lens one needs to find the focal length of the lens first. The focal length of the lens can be found using the lens formula,

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

Where $u$ is the object distance.

Step 3 - Finding the object's distance

The formula for magnification$\left(m\right)$ is

$m=\frac{v}{u}...\left(1\right)$

Magnification is the ratio of the height of the image to the height of the object.

So, another formula for magnification is given by

$m=\frac{h\text{'}}{h}...\left(2\right)$

Now by equating equation$\left(1\right)$and $\left(2\right)$we get

$\frac{v}{u}=\frac{h\text{'}}{h}$

In the question, it is given that the size of the image is the same as that of the object and the image is real and inverted, i.e. $h=-h\text{'}$

Substitite $-h\text{'}$for $h$and $25cm$ for $v$into the above equation to calculate $u$.

$$\begin{gathered} \frac{25cm}{u}=\frac{h\text{'}}{-h\text{'}} \\ \frac{25}{u}=-1 \\ u=-\frac{25}{1} \\ u=-25cm \end{gathered}$$

Step 4 - Finding the focal length of the lens.

Substitute the values into the lens formula to calculate the focal length of the lens.

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{f}=\frac{1}{25cm}-\frac{1}{(-25cm)} \\ \frac{1}{f}=\frac{1}{25}+\frac{1}{25} \\ \frac{1}{f}=\frac{2}{25} \\ f=\frac{25}{2} \\ f=12.5cm \end{gathered}$$

Step 5 - Finding the power of the lens.

Substitute the values into the formula for power.

$$\begin{gathered} P=\frac{1}{f} \\ P=\frac{1}{12.5}c{m}^{-1} \\ P=\frac{1}{12.5}{({10}^{-2}m)}^{-1} \\ P=\frac{100}{12.5}{m}^{-1} \\ P=8m^{-1} \\ P=8D \end{gathered}$$

(Note:- $1cm={10}^{-2}m$ and $1m^{-1}=1D$)

Hence, the needle should be placed at a distance of $25cm$ in front of the lens. The power of the lens is $8D$.