A thin diverging lens of focal length 25cm is placed 18cm to the right of a point source of blue light on the axis of the lens. Where is the image of the source? Is it a real or a virtual image? If you placed a sheet of paper at the location of the image, what would you see on the paper?

Consider that the point source of blue light is placed on the axis of a thin diverging lens is placed 18cm to the left of the lens. The focal length of the lens is 25cm.

Give the formula for magnification in terms of height and distance.

$\mathit{M}\mathbf{=}\mathbf{-}\frac{{\mathbf{d}}_{\mathbf{2}}}{{\mathbf{d}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{h}}_{\mathbf{2}}}{{\mathbf{h}}_{\mathbf{1}}}$ ……(1)

Give the lens equation.

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{2}}}$ ……(2)

Consider the Equation (2), Substitute the values of $f=-25\text{cm}$ and $d_1=-18\text{cm}$as follows,

$\frac{1}{\text{- 25cm}}=\frac{1}{\text{- 18cm}}+\frac{1}{d_2}$

Solve the above equation for $d_2$ as follows,

$$\begin{gathered} d_2=\frac{\left(\text{- 25cm}\right)\left(\text{- 18cm}\right)}{\text{- 18cm}-\left(-25\text{cm}\right)} \\ {d}_2=64.3\text{cm} \end{gathered}$$

The positive sign indicates that the image is real and is on the opposite side of the object.

Consider the Equation (1) and substitute the values as follows,

$$\begin{gathered} M=-\frac{d_2}{d_1} \\ M=-\frac{\left(-64.3cm\right)}{-18cm} \end{gathered}$$

$M=3.57$

Since the magnification of the lens is positive and greater than 1. So the image is upright and bigger than the object.

Therefore, if the sheet of paper is at location of the image, the real image and a magnified image will be formed on the screen.