The illuminance of a small -2 area changes from 900 lumen m^-2 to 400 lumen m^-2 when it is shifted along its normal by 10 cm. Assuming that it is illuminated by a point source placed on the normal, find the distance between the source and the area in the original position.

Explanation

To solve the problem, we will use the formula for illuminance (E) due to a point source:

\[ E = \frac{I \cdot \cos \theta}{r^2} \]

Where:
- \( E \) is the illuminance (in lumen/m²),
- \( I \) is the intensity of the point source,
- \( \theta \) is the angle between the normal to the surface and the line from the source to the surface,
- \( r \) is the distance from the source to the area.

### Step-by-Step Solution:

1. **Define Variables**:
- Let the original distance from the source to the area be \( X \) (in cm).
- The illuminance at this distance is given as \( E_A = 900 \, \text{lumen/m}^2 \).
- The illuminance after moving the area 10 cm further away is \( E_B = 400 \, \text{lumen/m}^2 \).
- The new distance from the source to the area after shifting is \( X + 10 \) cm.

2. **Set Up the Equations**:
- For the original position:
\[ E_A = \frac{I \cdot \cos \theta}{X^2} = 900 \]
- For the new position:
\[ E_B = \frac{I \cdot \cos \theta}{(X + 10)^2} = 400 \]

3. **Divide the Two Equations**:
- Dividing the equation for \( E_A \) by the equation for \( E_B \):
\[ \frac{E_A}{E_B} = \frac{(I \cdot \cos \theta) / X^2}{(I \cdot \cos \theta) / (X + 10)^2} \]
- This simplifies to:
\[ \frac{900}{400} = \frac{(X + 10)^2}{X^2} \]
- Simplifying the left side gives:
\[ \frac{9}{4} = \frac{(X + 10)^2}{X^2} \]

4. **Cross Multiply**:
- Cross multiplying gives:
\[ 9X^2 = 4(X + 10)^2 \]

5. **Expand the Right Side**:
- Expanding \( 4(X + 10)^2 \):
\[ 4(X^2 + 20X + 100) = 4X^2 + 80X + 400 \]
- So the equation becomes:
\[ 9X^2 = 4X^2 + 80X + 400 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 9X^2 - 4X^2 - 80X - 400 = 0 \]
- This simplifies to:
\[ 5X^2 - 80X - 400 = 0 \]

7. **Divide by 5**:
- Dividing the entire equation by 5:
\[ X^2 - 16X - 80 = 0 \]

8. **Using the Quadratic Formula**:
- The quadratic formula is given by:
\[ X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 1, b = -16, c = -80 \):
\[ X = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot (-80)}}{2 \cdot 1} \]
\[ = \frac{16 \pm \sqrt{256 + 320}}{2} \]
\[ = \frac{16 \pm \sqrt{576}}{2} \]
\[ = \frac{16 \pm 24}{2} \]

9. **Calculating the Roots**:
- This gives us two potential solutions:
\[ X = \frac{40}{2} = 20 \quad \text{(valid)} \]
\[ X = \frac{-8}{2} = -4 \quad \text{(not valid)} \]

10. **Conclusion**:
- Therefore, the distance between the source and the area in the original position is:
\[ X = 20 \, \text{cm} \]