In a slide show programme, the image on the screen has an area 900times that of the slide. If the distance between the slide and the screen is x times the distance between the slide and the projector lens, then

Options

A. `x=30`
B. `x=31`
C. `x=500`
D. `x=1//30`

Explanation

To solve the problem, we need to analyze the relationship between the areas of the slide and the image on the screen, the magnification, and the distances involved in the projection setup.

### Step-by-Step Solution:

1. **Understanding the Areas**:
- Let the area of the slide be \( A \).
- The area of the image on the screen is given as \( 900A \).

2. **Finding the Magnification of Area**:
- The magnification of area is defined as the ratio of the area of the image to the area of the object (slide).
- Therefore, the magnification of area \( M_a \) can be expressed as:
\[ M_a = \frac{\text{Area of Image}}{\text{Area of Slide}} = \frac{900A}{A} = 900 \]

3. **Finding Linear Magnification**:
- The linear magnification \( m \) is related to the magnification of area by the square root:
\[ m = \sqrt{M_a} = \sqrt{900} = 30 \]

4. **Defining Distances**:
- Let the distance between the slide and the projector lens be \( n \).
- The distance between the projector lens and the screen, based on the linear magnification, is:
\[ V = m \cdot n = 30n \]

5. **Calculating the Total Distance**:
- The total distance between the slide and the screen is the sum of the distance from the slide to the projector and from the projector to the screen:
\[ \text{Distance between slide and screen} = n + 30n = 31n \]

6. **Relating to the Given Ratio**:
- The problem states that the distance between the slide and the screen is \( x \) times the distance between the slide and the projector lens. Therefore:
\[ x \cdot n = 31n \]
- Dividing both sides by \( n \) (assuming \( n \neq 0 \)):
\[ x = 31 \]

### Final Answer:
The value of \( x \) is \( 31 \).