A lens is placed between a source of light and screen. It forms real image on screen for two different positions. If height of one image is 20 cm and the other is 80 cm, then the height of source of light (in cm) is_____.

Explanation

To solve the problem, we will use the concept of magnification in optics. The magnification (M) of a lens is given by the formula:

\[ M = \frac{h'}{h_0} \]

where:
- \( h' \) is the height of the image,
- \( h_0 \) is the height of the object (source of light).

In this case, we have two different image heights formed by the same lens at two different positions:

1. \( h'_1 = 20 \, \text{cm} \)
2. \( h'_2 = 80 \, \text{cm} \)

The relationship between the magnifications for the two positions of the lens can be expressed as:

\[ M_1 \times M_2 = 1 \]

This means:

\[ \left( \frac{h'_1}{h_0} \right) \times \left( \frac{h'_2}{h_0} \right) = 1 \]

Substituting the values of \( h'_1 \) and \( h'_2 \):

\[ \left( \frac{20}{h_0} \right) \times \left( \frac{80}{h_0} \right) = 1 \]

This simplifies to:

\[ \frac{20 \times 80}{h_0^2} = 1 \]

Calculating \( 20 \times 80 \):

\[ 20 \times 80 = 1600 \]

So we have:

\[ \frac{1600}{h_0^2} = 1 \]

Now, cross-multiplying gives:

\[ 1600 = h_0^2 \]

Taking the square root of both sides to find \( h_0 \):

\[ h_0 = \sqrt{1600} \]

Calculating the square root:

\[ h_0 = 40 \, \text{cm} \]

Thus, the height of the source of light is:

**Final Answer: \( 40 \, \text{cm} \)**