Question

A photographic slide is to the left of a lens. The lens projectsan image of the slide onto a wall 6.00 m to the right of the slide. The image is 80.0 times the size of the slide.

(a) How far is the slide from the lens?

(b) Is the image erect or inverted?

(c) What is the focal length of the lens?

(d) Is the lens converging or diverging?

Short answer

1. s=0.074 cm
2. The image is Inverted
3. The focal length of the lens is 0.073 cm.
4. The lens is converging.

Step-by-step solution

Given values.

d=6 m- the distance between the object and image M= 80 - the magnification of the lens

Required

a) It is necessary to determine the distance between the lens and a photographic slide.

b) lt is necessary to examine whether the image is inverted or not t is necessary to determine the lens focal length and to examine whether it is a converging or diverging lens.

Explanation.

The magnification of the lens that is a consequence of a certain lens can be determined based on the ratio between the dimensions between the object and its image or based on the distance from the focal point to the center of the lens

$\mathrm{M}=\frac{\mathrm{s}\text{'}}{\mathrm{s}}$ (1)

In the equation above, s' represents the distance between the object and its image while s is the distance from the object to the lens.

The focal length of a converging lens can be determined based on the known equation:

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}}$ (2)

Solving part (a) of the problem.

Using equation (1) we can determine the position of a photographic slide relative to the focal point s. Also, it is known that d = s + s'.

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{s}\text{'}}{\mathrm{s}} \\ \mathrm{s}\text{'}=\mathrm{Ms} \end{gathered}$$

Now, we can introduce it into the mentioned equation (d=s+ s').

$$\begin{gathered} \mathrm{d}=\mathrm{s}+\mathrm{s}\text{'} \\ =\mathrm{s}+\mathrm{Ms} \\ =\left(\mathrm{M}+1\right)\mathrm{s} \\ \mathrm{s}=\frac{\mathrm{d}}{\mathrm{M}+1} \\ =\frac{6\mathrm{m}}{80+1} \\ =0.074\mathrm{m} \end{gathered}$$

Based on the magnification M we can determine s'.

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{s}\text{'}}{\mathrm{s}} \\ \mathrm{s}\text{'}=\mathrm{M}.\mathrm{s} \\ =80.0.074\mathrm{m} \\ =5.926\mathrm{m} \end{gathered}$$

Solving parts (b) and (c) of the problem.

Based on the known quantities s and s' that is the known position of the image and object relative to the lens, we can determine its focal length. To do that, it is necessary to suppose the lens is a converging one. We will use the equation (2)

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}} \\ =\frac{\mathrm{s}\text{'}+\mathrm{s}}{\mathrm{s}.\mathrm{s}\text{'}} \\ \mathrm{f}=\frac{\mathrm{s}.\mathrm{s}}{\mathrm{s}+\mathrm{s}\text{'}} \\ =\frac{0.074\mathrm{m}.5.926}{5\mathrm{m}} \\ =0.073\mathrm{m} \end{gathered}$$

The previous confirms that it is a converging lens and we have determined its focal length.

Solving part (d) of the problem.

Since 2f>s>f then we can conclude that the image of an object is inverted, magnified, and inverted relative to the object The figure below illustrates this case. the lens is converging.