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Found in: Page 1043

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Figure 34-50a is an overhead view of two vertical plane mirrors with an object O placed between them. If you look into the mirrors, you see multiple images of O. You can find them by drawing the reflection in each mirror of the angular region between the mirrors, as is done in Fig. 34-50b for the left-hand mirror. Then draw the reflection of the reflection. Continue this on the left and on the right until the reflections meet or overlap at the rear of the mirrors. Then you can count the number of images of O. How many images of O would you see if ${\mathbit{\theta }}$ is (a) ${\mathbf{90}}{\mathbf{°}}$, (b) ${\mathbf{45}}{\mathbf{°}}$, and (c) ${\mathbf{60}}{\mathbf{°}}$? If ${\mathbit{\theta }}{\mathbf{=}}{\mathbf{120}}{\mathbf{°}}$, determine the (d) smallest and (e) largest number of images that can be seen, depending on your perspective and the location of O. (f) In each situation, draw the image locations and orientations as in Fig. 34-50b.

1. Reflected images of O from angle $90°$ are three.
2. Reflected images of O from angle $45°$ are seven.
3. Reflected images of O from angle $60°$ are five.
4. Smallest number of image from angle role="math" localid="1663017238465" $\theta =120°$ is one.
5. Largest images from angle $\theta =120°$ are three.
6. In each situation, the image locations and orientations are drawn.
See the step by step solution

## Step 1: The given data

Figure 34-50a is an overhead view of two vertical plane mirrors with an object O placed between them. Fig. 34-50b can be shown for the left-hand mirror.

## Step 2: Understanding the concept of multiple reflections

When two reflecting surfaces are kept in an adjacent manner to each other, they tend to reflect the light coming from the object kept at a distance from them resulting in a production of several images due to reflection. This case is defined as the case of multiple reflections. Here, the object when kept in front of a mirror, the reflected image is again reflected by the other mirror; that is image to mirror 1 is an object to mirror 2. These are generally used in cases where the image of the object is not clear and thus requires an additional mirror to create several images of the object when viewed from different angles.

Formula:

The number of images formed due to two plane mirrors kept at an angle $\theta$,

localid="1663017440181" $N=\frac{{360}^{0}}{\theta }-1,\frac{{360}^{0}}{\theta }\text{if even}\phantom{\rule{0ex}{0ex}}=\frac{{360}^{0}}{\theta },\text{if}\frac{{360}^{0}}{\theta }\text{is odd}\phantom{\rule{0ex}{0ex}}$ ...(i)

## Step 3:Calculation of reflected images for 900

(a)

Using the angle value $\theta =90°$ in equation (i), the number of images formed due to the plane mirror combination can be given as follows:

$N=\frac{{360}^{0}}{{90}^{0}}-1\left(\because \frac{{360}^{0}}{{90}^{0}}=4\text{is even.}\right)\phantom{\rule{0ex}{0ex}}=4-1\phantom{\rule{0ex}{0ex}}=3\phantom{\rule{0ex}{0ex}}$

Now, the positions and orientations of images can be given by the following the multiple reflection concept. That is, the object placed between them will first produce an image for each mirror at an equal distance from both the first and second mirror respectively. Now consider, when the ray of light coming from the object gets reflected by the first mirror then gets reflected by the second mirror ultimately produces a third image due to multiple reflection by the mirrors.

Thus, both ${I}_{1}$ and ${I}_{3}$ lies at a perpendicular distance to the mirror equally as the object is placed for each mirror respectively and ${I}_{3}$ is the virtualimage formed due to the parallel rays from the multiple reflections.

Hence, there are three reflected images.

## Step 4: Calculation of reflected images for 450

(b)

Using the angle value $\theta =45°$ in equation (i), the number of images formed due to the plane mirror combination can be given as follows:

$N=\frac{{360}^{0}}{{45}^{0}}-1\left(\because \frac{{360}^{0}}{{45}^{0}}=8\text{is even.}\right)\phantom{\rule{0ex}{0ex}}=8-1\phantom{\rule{0ex}{0ex}}=7\phantom{\rule{0ex}{0ex}}$

Now, the positions and orientations of images can be given by the following the multiple reflection concept. First, we get an image ${I}_{1}$ that is equally at a distance behind the first mirror that the object is placed in front of the first mirror. Again, this object and the image formed will have images ${I}_{2}$ and ${I}_{3}$ behind the second mirror as it is a plane mirror. Now, these images will reflect through the surfaces of the second mirror upon extension acting an object to them thus, here two more images ${I}_{4}=I$ and ${I}_{5}=I\text{'}$ are formed behind mirror 2 as shown. Now, further first mirror plane upon extension will reflect through the surfaces of the first mirror to produce equal two more images ${I}_{6}$ and ${I}_{7}$ that are twins to each other, thus there will be no more images after this.

The orientation and location are shown through the construction in the figure below. Summarizing, we find 1 + 2 + 2 + 2 = 7 images in this case.

Hence, there are seven reflected images.

## Step 5: Calculation of reflected images for 600

(c)

Using the angle value $\theta =60°$ in equation (i), the number of images formed due to the plane mirror combination can be given as follows:

$N=\frac{{360}^{0}}{{60}^{0}}-1\left(\because \frac{{360}^{0}}{{60}^{0}}=6\text{is even.}\right)\phantom{\rule{0ex}{0ex}}=6-1\phantom{\rule{0ex}{0ex}}=5\phantom{\rule{0ex}{0ex}}$

Now, the positions and orientations of images can be given by the following the multiple reflection concept. First, we get an image ${I}_{1}$ that is equally at a distance behind the first mirror that the object is placed in front of the first mirror. Again, this object and the image formed will have images ${I}_{2}$ and ${I}_{3}$ behind the second mirror as it is a plane mirror. Now, these images will reflect through the surfaces of the second mirror upon extension acting an object to them thus, here two more images ${I}_{4}=I$ and ${I}_{5}=I\text{'}$ are formed behind mirror 2. Now, the rays coming from these two images will be parallel and thus these images will be twin images and thus no further images will be produced

Summarizing, we find 1 + 2 + 2 = 5 images in this case.

Hence, there are five reflected images.

## Step 6: Calculation of the smallest number of images for 1200

(d)

Using the angle value $\theta =120°$ in equation (i), the number of images formed due to the plane mirror combination can be given as follows:

$N=\frac{{360}^{0}}{{120}^{0}}-1\left(\because \frac{{360}^{0}}{{120}^{0}}=3\text{is odd.}\right)\phantom{\rule{0ex}{0ex}}=3\phantom{\rule{0ex}{0ex}}$

Thus, it turns out in this case that the number of images that can be seen will vary from 1 to 3, depending on the locations of both the object and the observer. If the object distance and the image of the object are seen such that both are in collinear, then the minimum seen image is its reflection that is one.

Therefore, the smallest image from angle $\theta =120°$ will be, one.

## Step 7: Calculation of the largest number of images for 1200

(e)

Now, the positions and orientations of images can be given by the following the multiple reflection concept. First, we get an image ${I}_{1}$ that is equally at a distance behind the first mirror that the object is placed in front of the first mirror. Again, this object and the image formed will have images ${I}_{2}$ and ${I}_{3}$ behind the second mirror upon extension as it is a plane mirror. These images will turn out to be twin images, thus, there will be no further reflections. Thus, as per calculated in part (d), the number of images will be three.

Hence, the largest number of images is three.

## Step 8: Calculation of the image locations and orientations

(f)

For situation (b), the image locations and orientations can be given as follows:

For $\theta =90°$

For $\theta =45°$

For $\theta =60°$

For $\theta =120°$