An illuminated slide is held `68cm` from a screen. How far from the slide must a lens of focal length `11cm` be placed (between the slide and the screen) to form an image of the slide's picture on the screen ?

Answer

To solve the problem of determining how far from the slide a lens of focal length 11 cm must be placed to form an image of the slide's picture on the screen, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a slide (object) and a screen, with the distance between them being 68 cm.
- Let the distance from the slide to the lens be \( x \) cm. Then, the distance from the lens to the screen will be \( 68 - x \) cm.

2. **Identify the Lens Type**:
- Since we want to form a real image on the screen, we will use a convex lens. The focal length \( f \) of the lens is given as \( 11 \) cm.

3. **Apply the Lens Formula**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Here, \( u \) (object distance) is taken as negative according to the sign convention, so \( u = -x \) and \( v \) (image distance) is \( 68 - x \).

4. **Substitute Values into the Lens Formula**:
- Substitute \( u \) and \( v \) into the lens formula:
\[ \frac{1}{11} = \frac{1}{68 - x} - \frac{1}{-x} \]
- This simplifies to:
\[ \frac{1}{11} = \frac{1}{68 - x} + \frac{1}{x} \]

5. **Find a Common Denominator**:
- The common denominator for the right side is \( x(68 - x) \):
\[ \frac{68 - x + x}{x(68 - x)} = \frac{68}{x(68 - x)} \]
- Thus, we have:
\[ \frac{1}{11} = \frac{68}{x(68 - x)} \]

6. **Cross-Multiply**:
- Cross-multiplying gives:
\[ 68 \cdot 11 = x(68 - x) \]
- Simplifying this gives:
\[ 748 = 68x - x^2 \]
- Rearranging leads to:
\[ x^2 - 68x + 748 = 0 \]

7. **Solve the Quadratic Equation**:
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = -68 \), and \( c = 748 \).
- The discriminant \( D \) is:
\[ D = b^2 - 4ac = (-68)^2 - 4 \cdot 1 \cdot 748 = 4624 - 2992 = 1632 \]
- Thus, the roots are:
\[ x = \frac{68 \pm \sqrt{1632}}{2} \]

8. **Calculate the Values**:
- Calculate \( \sqrt{1632} \approx 40.4 \):
\[ x = \frac{68 \pm 40.4}{2} \]
- This gives two possible values for \( x \):
\[ x_1 = \frac{108.4}{2} \approx 54.2 \text{ cm} \]
\[ x_2 = \frac{27.6}{2} \approx 13.8 \text{ cm} \]

9. **Conclusion**:
- The lens can be placed approximately \( 54.2 \) cm or \( 13.8 \) cm from the slide to form an image on the screen.