Chapter 25, Problem 68GP

(a)

To determine

To Show: The exposer time must be four times longer.

Explanation of Solution

Given information:

Object height equals to the image height when the object is very close to camera

Formula used:

The magnification is given as,

  $m=-\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

Where, $h_i$ is the image height while $h_o$ is the object height.

Proof:

The focal point of the image will be $d_{i1}=f$ if object is very far away

According to the magnification,

  $m=-\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

Here, height is directly proportional to the distance

If the height of the object and image is same then the magnification becomes,

  $m=-\frac{h_i}{h_o}=-1$

And, so

  $\frac{d_{i2}}{d_{o2}}\Rightarrow d_{i2}=d_{o2}$

According to the Len’s equation

  $\frac{1}{d_{i2}}+\frac{1}{d_{o2}}=\frac{1}{f}\Rightarrow d_{i2}=2f\left(\because d_{i2}=d_{o2}\right)$

The required exposure time is directly proportional to the area of the image on the film

Therefore, the time ratio is

  $\frac{t_2}{t_1}={\left(\frac{d_{i2}}{d_{i1}}\right)}^2={\left(\frac{2f}{f}\right)}^2=4$

Conclusion:

The exposer time must be four times longer.

(b)

To determine

To Explain: The exposure time is increased by less than half a f-stop relative to the same object being a great distance away.

Explanation of Solution

Given:

Formula used:

The Len’s equation is,

  $\frac{1}{d_i}+\frac{1}{d_o}=\frac{1}{f}$

Where, f is the focal length, $d_i$ is the image distance and $d_o$ is the object distance.

Calculation:

For increased object distances, the lens equation is

  $\begin{array}{l}\frac{1}{d_{i3}}+\frac{1}{d_{o3}}=\frac{1}{f}\\ \frac{1}{4f}+\frac{1}{d_{i3}}=\frac{1}{f}\end{array}$

That gives the distance

  $\begin{array}{l}{d}_{i3}=\frac{4f}{3}\\ \frac{t_3}{t_1}={\left(\frac{4}{3}\right)}^2=1.78\end{array}$

  $\begin{array}{l}\frac{1}{d_{i4}}+\frac{1}{d_{o4}}=\frac{1}{f}\\ \frac{1}{5f}+\frac{1}{d_{i4}}=\frac{1}{f}\end{array}$

That gives,

  $\begin{array}{l}{d}_{i4}=\frac{5f}{4}\\ \frac{t_4}{t_1}={\left(\frac{5}{4}\right)}^2=1.56\end{array}$

These increased exposures are less than the minimal adjustment on a typical camera. Therefore, they are negligible.

Conclusion:

Yes, the exposure time is increased by less than half a f-stop relative to the same object