A lens placed between a candle and a fixed screen forms a real triply magnified image of the candle on the screen. When the lens is moved away form the candle by `0.8m` without changing the position of the candle, a real image one-third the size of the candle is formed on the screen. Determine the focal length of the lens.

Explanation

To solve the problem, we will use the lens formula and the magnification formula. Let's break it down step by step.

### Step 1: Understand the Problem
We have a lens that forms two images of a candle:
1. A real image that is three times the size of the candle (magnification = -3).
2. A real image that is one-third the size of the candle (magnification = -1/3) when the lens is moved away by 0.8 m.

### Step 2: Define Variables
Let:
- \( u_1 \) = initial object distance (distance from the candle to the lens)
- \( v_1 \) = image distance for the first case (when the image is triply magnified)
- \( u_2 \) = object distance for the second case (when the lens is moved away)
- \( v_2 \) = image distance for the second case (when the image is one-third the size)
- \( f \) = focal length of the lens

### Step 3: Use the Magnification Formula
The magnification \( m \) is given by:
\[ m = -\frac{v}{u} \]
For the first case (triply magnified image):
\[ -3 = -\frac{v_1}{u_1} \implies v_1 = 3u_1 \]

For the second case (one-third size image):
\[ -\frac{1}{3} = -\frac{v_2}{u_2} \implies v_2 = \frac{1}{3}u_2 \]

### Step 4: Relate Object Distances
Since the lens is moved away by 0.8 m, we have:
\[ u_2 = u_1 + 0.8 \]

### Step 5: Use the Lens Formula
The lens formula is:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

For the first case:
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} \]
Substituting \( v_1 = 3u_1 \):
\[ \frac{1}{f} = \frac{1}{3u_1} - \frac{1}{u_1} = \frac{1 - 3}{3u_1} = -\frac{2}{3u_1} \]
Thus,
\[ f = -\frac{3u_1}{2} \]

For the second case:
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \]
Substituting \( v_2 = \frac{1}{3}u_2 \):
\[ \frac{1}{f} = \frac{3}{u_2} - \frac{1}{u_2} = \frac{2}{u_2} \]
Thus,
\[ f = \frac{u_2}{2} \]

### Step 6: Set the Two Expressions for \( f \) Equal
From the two expressions for \( f \):
\[ -\frac{3u_1}{2} = \frac{u_2}{2} \]
Substituting \( u_2 = u_1 + 0.8 \):
\[ -\frac{3u_1}{2} = \frac{u_1 + 0.8}{2} \]
Multiplying through by 2 to eliminate the denominators:
\[ -3u_1 = u_1 + 0.8 \]
Rearranging gives:
\[ -4u_1 = 0.8 \implies u_1 = -0.2 \text{ m} \]

### Step 7: Substitute \( u_1 \) Back to Find \( f \)
Now substituting \( u_1 \) back into the expression for \( f \):
\[ f = -\frac{3(-0.2)}{2} = \frac{0.6}{2} = 0.3 \text{ m} \]

### Final Answer
The focal length of the lens is \( f = 0.3 \text{ m} \).

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