Chapter 17, Problem 92A

To determine

To explain: how the size of the image of ball changes as the ball rolls towards the concave mirror.

Answer to Problem 92A

The size of the image of the ball increases as the ball moves towards the concave mirror.

Explanation of Solution

Introduction:

Consider an object of height $h_{\text{o}}$ is placed at a distance $d_{\text{o}}$ from the front of spherical mirror, then an image of $h_{\text{i}}$ is formed at a distance $d_{\text{i}}$ from the mirror.

Show the expression for the magnification of the image as follows:

  $m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}$

The size of the image is larger than the size of the object, when the absolute value of the magnification of image is greater than 1.

The size of the image is smaller than the size of the object, when the absolute value of the magnification of image is lesser than 1.

Show the position of the ball as shown in Figure below.

  

Refer the Figure,

Consider the distance of the image from the mirror is denoted by $d_{\text{i}}$ .

Consider the distance of the object from the mirror is denoted by $d_{\text{o}}$ .

Position 1: ball is placed beyond the radius of curvature of the mirror.

The image will be formed between the radius of curvature $\left(\text{C}\right)$ and the focal length of the mirror $\left(\text{F}\right)$ .

Hence, the distance of the image from the mirror $\left(d_{\text{i}}\right)$ is lesser the distance of the object from the same mirror $\left(d_{\text{o}}\right)$ .

The absolute value of magnification of the image is lesser than 1 as $\left(d_{\text{i}}<d_{\text{o}}\right)$ .

Thus, the size of the image is smaller than the size of the object when the object is placed beyond the radius of curvature.

Position 2: ball is placed between the radius of curvatureand the focal length of the mirror.

The image will be formed beyond the radius of curvature $\left(\text{C}\right)$ of the mirror.

Hence, the distance of the image from the mirror $\left(d_{\text{i}}\right)$ is greater the distance of the object from the same mirror $\left(d_{\text{o}}\right)$ .

The absolute value of magnification of the image is greater than 1 as $\left(d_{\text{i}}>d_{\text{o}}\right)$ .

Thus, the size of the image is larger than the size of the object when the object is placed in between the radius of curvature and focal length of the mirror.

Position 3:ball is placed between the focal length and the mirror.

The absolute distance of the image from the mirror $\left|d_{\text{i}}\right|$ is greater the distance of the object from the same mirror $\left(d_{\text{o}}\right)$ .

The absolute value of magnification of the image is greater than 1 as $\left(\left|d_{\text{i}}\right|>d_{\text{o}}\right)$ .

Thus, the size of the image is larger than the size of the object when the object is placed in between the focal length and mirror.

Compare the size of the image of the object as its moves towards the mirror,

The size of the image of the object increases as it moves from position 1 to position 3.

Hence, the size of the image of the ball increases as the ball moves towards the concave mirror.

Conclusion:

Thus, the size of the image of the ball increases as the ball moves towards the concave mirror.