Question

An optical projector is used to project an image of a slide 50mm square on to a screen places 1.2 m from the lens. The image formed on the screen is 600mm square. Find the focal length of the projecting lens of the projector.
Note: convert all unit measurements to cm(centimeter)

Answer 1

To find the focal length of the projecting lens, we can use the lens formula and the magnification formula.

1. **Convert all measurements to centimeters**:
- Size of the slide: \( 50 \, \text{mm} = 5 \, \text{cm} \)
- Size of the image on the screen: \( 600 \, \text{mm} = 60 \, \text{cm} \)
- Distance from the lens to the screen: \( 1.2 \, \text{m} = 120 \, \text{cm} \)

2. **Calculate the magnification**:
\[
\text{Magnification} (M) = \frac{\text{Image size}}{\text{Object size}} = \frac{60 \, \text{cm}}{5 \, \text{cm}} = 12
\]

3. **Using the magnification formula**:
Magnification is also given by:
\[
M = -\frac{v}{u}
\]
where:
- \( v \) is the image distance (120 cm),
- \( u \) is the object distance.

We can rearrange this to find \( u \):
\[
12 = -\frac{120}{u} \implies u = -\frac{120}{12} = -10 \, \text{cm}
\]

The negative sign indicates that the object (the slide) is on the same side of the lens as the incoming light (which is conventional).

4. **Using the lens formula**:
The lens formula relates object distance, image distance, and the focal length:
\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\]
Plugging in the values:
\[
\frac{1}{f} = \frac{1}{120} + \frac{1}{-10}
\]
Find a common denominator (120):
\[
\frac{1}{f} = \frac{1}{120} - \frac{12}{120} = \frac{1 - 12}{120} = \frac{-11}{120}
\]
Therefore:
\[
f = -\frac{120}{11} \approx -10.91 \, \text{cm}
\]

However, in lens optics, we typically consider the focal length to be positive for a converging lens (which is usually the case for projectors). As it’s a projector lens, we take the absolute value:

Thus, the focal length of the projecting lens is approximately:
\[
\boxed{10.91 \, \text{cm}}
\]