An object is `5.0 m` to the left of a flat screen. A converging lens for which the focal length is `f=0.8 m` is placed between object and screen.
(a) Show that two lens positions exist that form images on the screen and deremine how far these positions are from the object?
(b) How do the two images differ from each other?

Explanation

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

### Step-by-Step Solution

#### Step 1: Understand the Setup
We have:
- An object located 5.0 m to the left of a flat screen.
- A converging lens with a focal length \( f = 0.8 \, \text{m} \).
- The distance from the object to the screen is \( 5.0 \, \text{m} \).

Let \( u \) be the distance from the lens to the object (which will be negative as per the sign convention), and \( v \) be the distance from the lens to the image on the screen. Since the total distance from the object to the screen is 5 m, we have:
\[ v = 5 - u \]

#### Step 2: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( v = 5 - u \) into the lens formula:
\[ \frac{1}{0.8} = \frac{1}{5 - u} - \frac{1}{u} \]

#### Step 3: Rearrange the Equation
Rearranging gives:
\[ \frac{1}{5 - u} + \frac{1}{u} = \frac{1}{0.8} \]
This can be rewritten as:
\[ \frac{u + (5 - u)}{u(5 - u)} = \frac{1}{0.8} \]
\[ \frac{5}{u(5 - u)} = \frac{1}{0.8} \]

#### Step 4: Cross Multiply and Simplify
Cross multiplying gives:
\[ 5 \cdot 0.8 = u(5 - u) \]
\[ 4 = 5u - u^2 \]
Rearranging leads to:
\[ u^2 - 5u + 4 = 0 \]

#### Step 5: Solve the Quadratic Equation
We can factor the quadratic equation:
\[ (u - 4)(u - 1) = 0 \]
Thus, the solutions for \( u \) are:
\[ u = 4 \, \text{m} \quad \text{and} \quad u = 1 \, \text{m} \]

#### Step 6: Calculate the Positions of the Lens
The distances from the object to the lens are:
- For \( u = 4 \, \text{m} \): The lens is \( 4 \, \text{m} \) from the object.
- For \( u = 1 \, \text{m} \): The lens is \( 1 \, \text{m} \) from the object.

### Part (b): Differences Between the Two Images
#### Step 7: Calculate the Magnification
The magnification \( m \) is given by:
\[ m = -\frac{v}{u} \]

- For \( u = 4 \, \text{m} \):
\[ v = 5 - 4 = 1 \, \text{m} \]
\[ m = -\frac{1}{4} = -0.25 \]

- For \( u = 1 \, \text{m} \):
\[ v = 5 - 1 = 4 \, \text{m} \]
\[ m = -\frac{4}{1} = -4 \]

#### Step 8: Interpret the Results
- The first image (for \( u = 4 \, \text{m} \)) is smaller than the object (magnification of -0.25).
- The second image (for \( u = 1 \, \text{m} \)) is larger than the object (magnification of -4).

### Final Answers
(a) The two lens positions are \( 4 \, \text{m} \) and \( 1 \, \text{m} \) from the object.
(b) The images differ in size: one is smaller and inverted, and the other is larger and inverted.