A thin converging lens of focal length 20cm is located at the origin and its axis lies on the x axis. A point source of red light is placed at location $<-25,1,0>\mathbf{\text{cm}}$. 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 20cm is placed at the origin and its axis lies on the x axis. A point source of red light is placed at location .

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

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

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

Give the lens equation.

$\mathit{M}\mathbf{=}\frac{{\mathbf{d}}_{\mathbf{2}}}{{\mathbf{d}}_{\mathbf{1}}}$ ……(3)

Consider that the converging lens is at origin and the axis lies on the x axis. The point source of red light is placed at the location . The x coordinate represents the object distance and the y coordinate represents the height of the object.

In Equation (3) ,substitute $f=20\text{cm}$ and $d_1=25\text{cm}$ as follows,

$\frac{1}{20\text{cm}}=\frac{1}{25\text{cm}}+1d_2$

Solve the above equation for $d_2$ as follows,

$$\begin{gathered} d_2=\frac{1}{20\text{cm}}-\frac{1}{25\text{cm}} \\ {d}_2=100\text{cm} \end{gathered}$$

Since the image distance is positive, the image is real.

The magnification formula is,

$-\frac{d_2}{d_1}=\frac{h_2}{h_1}$

Solve the above equation for $h_2$ as follows,

$$\begin{gathered} h_2=-\frac{100\text{cm}}{25\text{cm}}\left(1\text{cm}\right) \\ {h}_2=-4\text{cm} \end{gathered}$$

The negative sign indicates that the image is inverted. Since $h_2>h_1$, the image is larger than the object.

Therefore, the final position is .