Question

Figure 34-46a shows the basic structure of an old film camera. A lens can be moved forward or back to produce an image on film at the back of the camera. For a certain camera, with the distance i between the lens and the film set at f = 5.00 cm, parallel light rays from a very distant object O converge to a point image on the film, as shown. The object is now brought closer, to a distance of p = 100 cm, and the lens–film distance is adjusted so that an inverted real image forms on the film (Fig. 34-46b). (a) What is the lens–film distance i now? (b) By how much was distance i changed?

Short answer

1. The lens-film distanceis 5.3 cm.
2. The change in lens-film distance i is 0.30 cm.

Step-by-step solution

Step 1: Given data

• Focal length,$f=5.0\text{cm}$ .

• Object distance, $p=100\text{cm}$.

Determining the concept

Using the lens formula, calculate the lens-film distance, which is image distance i. Using this, the change in lens-film distance can be calculated.

Formulae are as follows:

$\frac{1}{f}=\left(\frac{1}{p}+\frac{1}{i}\right)$

Here, p is the pole, and i is the image distance, f is the focal length.

(a) Determining the lens-film distance i

According to equation 34-4, find the image distance as,

$\frac{1}{f}=\left(\frac{1}{p}+\frac{1}{i}\right)$

Rearranging this formula gives,

$\frac{1}{i}=\frac{1}{f}-\frac{1}{p}$

So,

$i={\left(\frac{1}{f}-\frac{1}{p}\right)}^{-1}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}i& =& {\left(\frac{1}{5}-\frac{1}{100}\right)}^{-1}\\ & =& \frac{100\times 5}{100-5}\\ & =& \frac{500}{95}\\ & =& 5.26\text{cm}\\ & & \end{array}$

So,

$i\approx 5.3\text{cm}$

Therefore, the lens-film distance is 5.3 cm.

(b) Determine the change in lens-film distance i

The change in the lens-film distance:

Change in lens-film distance is,

$\begin{array}{rcl}\mathrm{Change}& =& i-f\\ & =& 5.3-5.0\\ & =& 0.30\text{cm}\\ & & \end{array}$

Therefore, the change in lens-film distance i is 0.30 cm.