Question

A point source of green light is placed on the axis of a thin converging lens, 7cm 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.

Short answer

The final position is 23.33cm. It is a virtual image. The virtual image does not appear on paper.

Step-by-step solution

Explain the given information

Consider that the point source of green light is placed on the axis of a thin converging lens 7cm to the left of the lens. The focal length of the lens is 10cm.

Give the Formula used.

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)

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 the Equation (2), Substitute the values of $f=10\text{cm}$ and $d_1=7\text{cm}$as follows,

$\frac{1}{10\text{cm}}=\frac{1}{7\text{cm}}+\frac{1}{d_2}$

Solve the above equation for $d_2$ as follows,

$$\begin{gathered} d_2=\frac{\left(10\text{cm}\right)\left(7\text{cm}\right)}{7\text{cm}-10\text{cm}} \\ {d}_2=-23.33\text{cm} \end{gathered}$$

The negative sign indicates that the image is virtual and is on the same side of the object.

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

$$\begin{gathered} M=-\frac{d_2}{d_1} \\ M=-\frac{\left(-23.33\text{cm}\right)}{7\text{cm}} \\ M=3.33 \end{gathered}$$

Since the magnification of the lens is positive and greater than 1. So the image is virtual.

Therefore, if the sheet of paper is at location of the image, the image can’t be seen on the screen since the virtual image will not appear on the screen.