Chapter 9, Problem 8C

(a)

To determine

Distance of the film from the lens.

Explanation of Solution

Given:

Focal length of the lens, $f=50\text{mm}$

Distance of the object, $s=350\text{mm}$

Formula used:

The formula of the focal distance of the lens is given as

  $\frac{1}{f}=\frac{1}{s}+\frac{1}{p}$

Here,

  $f$ is the focal length of the lens

  $s$ is the distance of object from the lens

  $p$ is the distance of image form from the lens

Calculation:

The formula of the focal distance of the lens is given as

  $\frac{1}{f}=\frac{1}{s}+\frac{1}{p}$

Plugging the values in the above equation

  $\begin{array}{c}\frac{1}{50}=\frac{1}{350}+\frac{1}{p}\\ p=58.33\text{mm}\end{array}$

Conclusion:

Distance of the film from the lens is $58.33\text{mm}$ .

(b)

To determine

The magnification and the height of the image.

Explanation of Solution

Given:

Focal length of the lens, $f=50\text{mm}$

Distance of the object, $s=350\text{mm}$

Length of the object, $O=100\text{mm}$

Formula used:

Magnification of the lens is calculated as

  $M=\frac{-p}{s}=\frac{\text{image}\text{height}}{\text{object}\text{height}}$

Here,

  $s$ is the distance of object from the lens

  $p$ is the distance of image form from the lens

Calculation:

Magnification of the lens is calculated as

  $M=\frac{-p}{s}=\frac{\text{image}\text{height}}{\text{object}\text{height}}$

Plugging the values in the above equation

  $\begin{array}{c}M=\frac{-58.33}{350}\\ =-0.167\end{array}$

The height of the image is calculated as

  $M=\frac{-p}{s}=\frac{\text{image}\text{height}}{\text{object}\text{height}}$

Plugging the values in the above equation

  $\begin{array}{c}-0.167=\frac{I}{100}\\ I=-16.7\text{mm}\end{array}$

Conclusion:

The magnification and the height of the image are $-0.167$ and $-16.7\text{mm}$ respectively.

(c)

To determine

The lens-to-film distance, magnification and the height of the object.

Explanation of Solution

Given:

Focal length of the lens, $f=50\text{mm}$

Distance of the object, $s=3500\text{mm}$

Formula used:

The formula of the focal distance of the lens is given as

  $\frac{1}{f}=\frac{1}{s}+\frac{1}{p}$

Here,

  $f$ is the focal length of the lens

  $s$ is the distance of object from the lens

  $p$ is the distance of image form from the lens

Magnification of the lens is calculated as

  $M=\frac{-p}{s}=\frac{\text{image}\text{height}}{\text{object}\text{height}}$

Here,

  $s$ is the distance of object from the lens

  $p$ is the distance of image form from the lens

Calculation:

The formula of the focal distance of the lens is given as

  $\frac{1}{f}=\frac{1}{s}+\frac{1}{p}$

Plugging the values in the above equation

  $\begin{array}{c}\frac{1}{50}=\frac{1}{3500}+\frac{1}{p}\\ p=50.72\text{mm}\end{array}$

Magnification of the lens is calculated as

  $M=\frac{-p}{s}=\frac{\text{image}\text{height}}{\text{object}\text{height}}$

Plugging the values in the above equation

  $\begin{array}{c}M=\frac{-50.72}{3500}\\ =-0.0144\end{array}$

The height of the image is calculated as

  $M=\frac{-p}{s}=\frac{\text{image}\text{height}}{\text{object}\text{height}}$

Plugging the values in the above equation

  $\begin{array}{c}-0.0144=\frac{I}{100}\\ I=-1.44\text{mm}\end{array}$

Conclusion:

Distance of the image from the lens is $50.72\text{mm}$ , magnification of the lens is $-0.0144$ , and the height of the image is $-1.44\text{mm}$ .