In Fig. 34-52, an object is placed in front of a converging lens at a distance equal to twice the focal length ${\mathit{f}}_{\mathbf{1}}$of the lens. On the other side of the lens is a concave mirror of focal length${\mathit{f}}_{\mathbf{2}}$separated from the lens by a distance $\mathbf{2}\mathbf{(}{\mathit{f}}_{\mathbf{1}}\mathbf{+}{\mathit{f}}_{\mathbf{2}}\mathbf{)}$. Light from the object passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image of the object. What are (a) the distance between the lens and that final image and (b) the overall lateral magnification M of the object? Is the image (c) real or virtual (if it is virtual, it requires someone looking through the lens toward the mirror), (d) to the left or right of the lens, and (e) inverted or non-inverted relative to the object?

Figure 34-52 shows an object in front of a converging lens with object distance$p_1=2f_1$ and there is a concave mirror at a distance $2\left(f_1+f_2\right)$from the lens.

When an object is placed in front of a converging lens at a distance, the real image will form to the right of the lens, i.e., at an image distance in front of the mirror as shown in figures 34-52. To find the distance between the lens and the final image, we will use the lens equation. Also, magnification can be calculated by taking the ratio of the focal lengths.

Formula:

The lens formula,

$\frac{\mathbf{1}}{\mathit{f}}\mathbf{=}\frac{\mathbf{1}}{\mathit{p}}\mathbf{+}\frac{\mathbf{1}}{\mathit{i}}$ ...(i)

The magnification formula of the lens,

$\mathit{m}\mathbf{=}\mathbf{-}\frac{\mathit{i}}{\mathit{p}}$ ...(ii)

The overall lateral magnification of two lenses,

$\mathit{m}\mathbf{=}\mathbf{\prod }_{\mathbf{i}}^{\mathbf{n}}{\mathit{m}}_{\mathbf{i}}$ ...(iii)

(a)

Firstly, the lens forms the real image of the object located at a distance. Now, using equation (i), the equation of the image distance can be given as follows:

$$\begin{gathered} \frac{1}{i_1}=\frac{1}{f_1}-\frac{1}{p_1} \\ \frac{1}{i_1}=\frac{1}{f_1}-\frac{1}{2f_1}\left(\because p_1=2f_1\right) \\ {i}_1=2f_1 \end{gathered}$$

Image is formed to the right of the lens.

Also, on the other side of the lens is a concave mirror of focal length $f_2$separated from the lens by a distance$2\left(f_1+f_2\right)$.

i.e., the object distance for the concave mirror is given by:

$$\begin{gathered} p_2=2\left(f_1+f_2\right)-2f_1 \\ =2f_2 \end{gathered}$$

The object is in front of the mirror.

Now,the light from the object passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image$i_1\text{'}$of the object that is given using equation (i) as follows:

$\frac{1}{i_1\text{'}}=\frac{1}{f_1}-\frac{1}{p_1\text{'}}$

Where, $p_1\text{'}$is the object distance =$2f_1$

Thus, the above equation becomes:

$$\begin{gathered} \frac{1}{i_1^{\text{'}}}=\frac{1}{f_1}-\frac{1}{2f_1} \\ \frac{1}{i_1^{\text{'}}}=\frac{1}{2f_1} \\ {i}_1^{\text{'}}=2f_1 \end{gathered}$$

This means the distance between the lens and final image isthe same as the location of the original object.

(b)

The overall magnification can be found by using equations (ii) and (iii) as follows:

$$\begin{gathered} m=\left(-\frac{i_1}{p_1}\right)\left(-\frac{i_2}{p_2}\right)\left(-\frac{i_1\text{'}}{p_1\text{'}}\right) \\ =\left(-\frac{2f_1}{2f_1}\right)\left(-\frac{2f_2}{2f_2}\right)\left(-\frac{2f_1}{2f_1}\right) \\ =\left(-1\right)\left(-1\right)\left(-1\right) \\ =-1.0 \end{gathered}$$

Hence, the value of magnification factor is localid="1663060570851" $-1.0$.

(c)

The final image is real (R) as it is present in front of the mirror.

Hence, the final image is real (R).

(d)

The final image is at a distance$i_1\text{'}$ to the left of the lens.

(e)

The final image is inverted as shown inthefigure below.

Hence, the image is inverted.