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

Answer 1

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

### Given:
- Size of the object (slide): \( h_0 = 50 \, \text{mm} = 0.05 \, \text{m} \)
- Size of the image on the screen: \( h_i = 600 \, \text{mm} = 0.6 \, \text{m} \)
- Distance from the lens to the screen (image distance): \( v = 1.2 \, \text{m} \)

### Step 1: Calculate Magnification
The magnification \( M \) is defined as the ratio of the height of the image to the height of the object. It is also equal to the ratio of the image distance to the object distance:

\[
M = \frac{h_i}{h_0} = \frac{v}{u}
\]

Where \( u \) is the object distance.

Calculating the magnification:

\[
M = \frac{0.6 \, \text{m}}{0.05 \, \text{m}} = 12
\]

### Step 2: Relate Object Distance and Image Distance Using Magnification
From the magnification equation, we have:

\[
M = \frac{v}{u} \implies u = \frac{v}{M}
\]

Substituting values:

\[
u = \frac{1.2 \, \text{m}}{12} = 0.1 \, \text{m}
\]

### Step 3: Use the Lens Formula
The lens formula relates the object distance \( u \), the image distance \( v \), and the focal length \( f \):

\[
\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
\]

Substituting the values we found for \( u \) and \( v \):

\[
\frac{1}{f} = \frac{1}{0.1} + \frac{1}{1.2}
\]

Calculating the terms:

\[
\frac{1}{0.1} = 10 \quad \text{and} \quad \frac{1}{1.2} \approx 0.8333
\]

Thus,

\[
\frac{1}{f} = 10 + 0.8333 \approx 10.8333
\]

### Step 4: Find the Focal Length
Now, taking the reciprocal to find \( f \):

\[
f \approx \frac{1}{10.8333} \approx 0.0922 \, \text{m} = 92.2 \, \text{mm}
\]

So, the focal length of the projecting lens is approximately:

\[
\boxed{92.2 \, \text{mm}}
\]