Q5E - University Physics with Modern Physics

Question

An object 0.6 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

Step-by-Step Solution

Answer

1. The diagram shows that the image is real and inverted. At the same time, it is magnified than the object
2. $s\text{'}=33cm$
   
3. $\left|y\text{'}\right|=1.2cm$
   

Step 1: Draw the ray diagram

To draw the principal rays, we have three principal rays coloured as shown in the figure below. The diagram shows that the image is real and inverted.

                            

Step 2: Calculate the position of the image

Given that the radius of curvature is $R=22cm$ and $S=16.5cm$.

The ratio between the distance of object and the distance of image is called lateral magnification and is given by

$m=‐\frac{s\text{'}}{s}=‐1$

The relation between the object’s distance and the image distance for the spherical mirrors is given by

$\frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}}=\frac{1}{\mathrm{f}}$

Now, the focal length is half the radius of curvature.

$$\begin{gathered} \frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}}=\frac{1}{\mathrm{f}} \\ \mathrm{s}\text{'}=\frac{\mathrm{sf}}{\mathrm{s}‐\mathrm{f}} \\ \mathrm{s}\text{'}=\frac{\mathrm{s}{\displaystyle \frac{\mathrm{R}}{2}}}{\mathrm{s}‐{\displaystyle \frac{\mathrm{R}}{2}}} \\ \mathrm{s}\text{'}=\frac{\mathrm{sR}}{2\mathrm{s}‐\mathrm{R}} \end{gathered}$$

Put the values in the equation and find the distance between the image and mirror.

$$\begin{gathered} \mathrm{s}\text{'}=\frac{22\times 16.5}{2\left(16.5\right)‐22} \\ \mathrm{s}\text{'}=33\mathrm{cm} \end{gathered}$$

Positive sign indicates that the image is to the left of the mirror.

Step 3: Calculate the height of the image

Calculate the lateral magnification

$\mathrm{m}=‐\frac{\mathrm{s}\text{'}}{\mathrm{s}}=‐\frac{33}{16.5}=‐2$

Also, the lateral magnification is given by $\mathrm{m}=‐\frac{\mathrm{y}\text{'}}{\mathrm{y}}$ . So,

$$\begin{gathered} \left|\mathrm{y}\text{'}\right|=\left|\mathrm{my}\right| \\ \left|\mathrm{y}\text{'}\right|=\left|‐2\times 0.6\right| \\ \left|\mathrm{y}\text{'}\right|=1.2\mathrm{cm} \end{gathered}$$