Question

A movie camera with a (single) lens of focal length $\mathbf{75}\mathbf{}\mathbf{mm}$takesa picture ofa person standing $\mathbf{27}\mathbf{}\mathbf{m}$away. If the person is$\mathbf{180}\mathbf{}\mathbf{cm}$tall, what is the height of the image on the film?

Short answer

The height of the image on the film is $0.5\mathrm{cm}$.

Step-by-step solution

Listing the given quantities

Focal length

$$\begin{gathered} \mathrm{f}=75\mathrm{mm} \\ =7.5\mathrm{cm} \end{gathered}$$

Object distance

$$\begin{gathered} \mathrm{p}=27\mathrm{m} \\ =2700\mathrm{cm} \end{gathered}$$

Height of the person$\mathrm{H}=180\mathrm{cm}$

Understanding the concepts of optics

The focal length and the object distance is given in the problem. By using the lens equation, we can find the image distance. Using the relation between the magnification, the image distance and the object distance, we can find the magnification. Using the height of the object given in the problem, we can find the height of the image.

Formula:

$$\begin{gathered} \frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{p}} \\ \mathbf{m}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\mathbf{p}} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{Image}\mathbf{}\mathbf{height}\mathbf{(}\mathbf{h}\mathbf{\text{'}}\mathbf{)}}{\mathbf{Original}\mathbf{}\mathbf{height}\mathbf{}\mathbf{\left(}\mathbf{H}\mathbf{\right)}} \end{gathered}$$

Calculations of the height of the image on the film

We have,

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{i}}+\frac{1}{\mathrm{p}}$

Rearranging for,

$$\begin{gathered} \frac{1}{\mathrm{i}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{p}} \\ \mathrm{i}=\frac{1}{\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{p}}} \\ =\frac{1}{\frac{1}{7.5}-\frac{1}{2700}} \\ =7.5\mathrm{cm} \end{gathered}$$

We have,

$$\begin{gathered} \mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}} \\ \mathrm{m}=\frac{\mathrm{Image}\mathrm{height}\left(\mathrm{h}\text{'}\right)}{\mathrm{Original}\mathrm{height}\left(\mathrm{H}\right)} \\ -\frac{\mathrm{i}}{\mathrm{p}}=\frac{\mathrm{Image}\mathrm{height}\left(\mathrm{h}\text{'}\right)}{\mathrm{Original}\mathrm{height}\left(\mathrm{H}\right)} \end{gathered}$$

$\begin{array}{rcl}\mathrm{Image}\mathrm{height}\left(\mathrm{h}\text{'}\right)& =& -\frac{\mathrm{i}}{\mathrm{p}}\times \mathrm{Original}\mathrm{height}\left(\mathrm{H}\right)\\ & =& -\frac{7.5}{2700}\times 180\\ & =& 0.5\mathrm{cm}\\ & & \end{array}$

The height of the image on the film is $=0.5\mathrm{cm}$.