A thin converging lens of focal length 20m is located at the origin and its axis lies on the x axis. A point source of red light is placed at location $\mathbf{<}\mathbf{-}\mathbf{14}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{\text{cm}}$. Where is the location of the image of the source? Is it a real or a virtual image? It helps to draw a diagram and trace the “easy” rays.

Consider that the focal length of a thin converging lens is 20mis located at the origin and tis axis lies on the x-axis. A point source is placed at location.

Give the formula for magnification considering $y\text{'}$ is the size of the image and y is the size of the object.

$\mathit{M}\mathbf{=}\frac{\mathbf{y}\mathbf{\text{'}}}{\mathbf{y}}$ ……(1)

Give the formula for magnification considering d is the object distance and $d\text{'}$ is the image distance from the center of the lens.

localid="1668581126450" $\mathit{M}\mathbf{=}\mathbf{-}\frac{\mathbf{d}\mathbf{\text{'}}}{\mathbf{d}}$ ……(2)

Give the lens equation considering d is the object distance and d' is the image distance from the center of the lens.

localid="1668581138594" $\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{d}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{d}\mathbf{\text{'}}}$

Consider the following sign convention of the lens equation and the magnification equation.

• Converging lens has the positive focal length and a diverging lens has the negative focal length.
• If the image is real and on the opposite side of the lens, then the distance between the image and the lens will be positive.
• If the image virtual and on same side of the lens, then the distance between the image and the lens will be negative.
• If the image is upright then it will have positive height and if the image is inverted it will have negative height.

Consider that the position of the object is and so, the object is 1cm away from the centre of the lens along y-axis, $d=1\text{cm}$ and the height of the object is $y=14\text{cm}$.

Given is the converging lens so, the focal length will be positive and the lens equation is as follows,

$\frac{1}{f}=\frac{1}{d}+\frac{1}{d\text{'}}$

$\frac{1}{20}=\frac{1}{1}+\frac{1}{d\text{'}}$

Solve the above equation for $d\text{'}$ as follows,

$$\begin{gathered} d\text{'}=-\frac{20}{19} \\ d\text{'}=-1.05\text{cm} \end{gathered}$$

The $d\text{'}$ is negative, so the image is virtual and is on the same side of the lens.

Consider the magnification equation,

$-\frac{d\text{'}}{d}=\frac{y\text{'}}{y}$

Substitute the values,

$-\frac{-1.05}{1}=\frac{y\text{'}}{14}$

Solve the above equation for $y\text{'}$,

$y\text{'}=14.7cm$

Hence, the height of the image is positive, this indicates that the image is upright and is virtual.

Therefore, the location of the image of the source is 1.05cm and it is a virtual image.