A point source of green light is placed on the axis of a thin converging lens, 15cm to the left of the lens. The lens has a focal length of 10cm. Where is the location of 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 green light is placed on the axis of a thin converging lens 15cm to the left of the lens. The focal length of the lens is 10cm.

Give the lens equation.

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

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

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

Consider the Equation (1), and rewrite it to solve it for $d_2$

$$\begin{gathered} \frac{1}{f}=\frac{1}{d_1}+\frac{1}{d_2} \\ {d}_2=\frac{f{d}_1}{d_1-f} \end{gathered}$$

Substitute the values of $f=10\text{cm}$ and $d_1=15\text{cm}$and Solve the above equation for $d_2$ as follows,

$$\begin{gathered} d_2=\frac{\left(10\text{cm}\right)\left(\text{15cm}\right)}{\text{15cm}-10\text{cm}} \\ {d}_2=30\text{cm} \end{gathered}$$

AS per the above result, the image is real.

Consider the Equation (2) and substitute the values as follows,

$$\begin{gathered} M=-\frac{\left(-30\text{cm}\right)}{\text{15cm}} \\ M=-2 \end{gathered}$$

Since the magnification of the lens is negative and greater than 1. So the image is inverted and the image will be taller in appearance.