Chapter 34: Q106P (page 1044)

In Fig. 34-52, an object is placed in front of a converging lens at a distance equal to twice the focal length $f_{1}$ of the lens. On the other side of the lens is a concave mirror of focal length $f_{2}$ separated from the lens by a distance $2 (f_{1} + f_{2})$ . 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?

Short Answer

Expert verified

The distance between the lens and final image is thesame as the location of the original object.
Magnification M of the object is $- 1.0$ .
The final image is real.
The image is at the left of the lens at distance $i_{1}$ .
The image is inverted.

Step by step solution

The given data

Figure 34-52 shows an object in front of a converging lens with object distance $p_{1} = 2 f_{1}$ and there is a concave mirror at a distance $2 (f_{1} + f_{2})$ from the lens.

Understanding the concept of lens-mirror system

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,

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

The magnification formula of the lens,

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

The overall lateral magnification of two lenses,

$m = \prod_{i}^{n} m_{i}$ ...(iii)

Step 3:Calculation of the distance between lens and final image

(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:

$\frac{1}{i_{1}} = \frac{1}{f_{1}} - \frac{1}{p_{1}} \frac{1}{i_{1}} = \frac{1}{f_{1}} - \frac{1}{2 f_{1}} (∵ p_{1} = 2 f_{1}) i_{1} = 2 f_{1}$

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 $2 (f_{1} + f_{2})$ .

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

$p_{2} = 2 (f_{1} + f_{2}) - 2 f_{1} = 2 f_{2}$

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 $i_{1}'$ of the object that is given using equation (i) as follows:

$\frac{1}{i_{1}'} = \frac{1}{f_{1}} - \frac{1}{p_{1}'}$

Where, $p_{1}'$ is the object distance = $2 f_{1}$

Thus, the above equation becomes:

$\frac{1}{i_{1}^{'}} = \frac{1}{f_{1}} - \frac{1}{2 f_{1}} \frac{1}{i_{1}^{'}} = \frac{1}{2 f_{1}} i_{1}^{'} = 2 f_{1}$

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

Calculation of the magnification of the object

(b)

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

$m = (- \frac{i_{1}}{p_{1}}) (- \frac{i_{2}}{p_{2}}) (- \frac{i_{1}'}{p_{1}'}) = (- \frac{2 f_{1}}{2 f_{1}}) (- \frac{2 f_{2}}{2 f_{2}}) (- \frac{2 f_{1}}{2 f_{1}}) = (- 1) (- 1) (- 1) = - 1.0$