Sunlight of intensity `1.3kW m^(–2)` is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kW `m^(–2)`, at a distance 22 cm from the lens on the other side is __________.

Answer

To solve the problem, we need to find the average intensity of light at a distance of 22 cm from the lens, given that sunlight of intensity 1.3 kW/m² is incident normally on a thin convex lens of focal length 20 cm.

### Step-by-Step Solution:

**Step 1: Understand the setup**
- We have a thin convex lens with a focal length (f) of 20 cm.
- Sunlight is incident normally on the lens with an intensity (I₁) of 1.3 kW/m².

**Step 2: Identify the distances**
- The distance from the lens to the point where we want to find the intensity (d) is 22 cm.

**Step 3: Use the lens formula**
- Since the lens is thin and the aperture size is much smaller than its focal length, we can use the lens formula to find the relationship between the distances.
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) = focal length (20 cm)
- \( v \) = image distance (unknown)
- \( u \) = object distance (infinity for sunlight)

- For sunlight, we can assume \( u \) is very large, thus \( \frac{1}{u} \approx 0 \).
- Therefore, \( \frac{1}{f} = \frac{1}{v} \) implies \( v = f = 20 \) cm.

**Step 4: Calculate the area ratio**
- The area of the light incident on the lens (A₁) can be calculated based on the radius of the lens.
- The area where we want to find the intensity (A₂) is at a distance of 22 cm from the lens.
- The ratio of the areas (A₁/A₂) can be determined from the distances:
\[ \frac{A_1}{A_2} = \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{f}{d}\right)^2 \]
where \( R_1 = 20 \) cm and \( R_2 = 22 \) cm.

\[ \frac{A_1}{A_2} = \left(\frac{20}{22}\right)^2 = \left(\frac{20}{22}\right)^2 = \left(\frac{10}{11}\right)^2 = \frac{100}{121} \]

**Step 5: Apply the conservation of energy**
- The conservation of energy states that the product of intensity and area remains constant:
\[ I_1 A_1 = I_2 A_2 \]
- Rearranging gives:
\[ I_2 = I_1 \frac{A_1}{A_2} \]
- Substituting the known values:
\[ I_2 = 1.3 \, \text{kW/m}^2 \times \frac{100}{121} \]

**Step 6: Calculate the intensity at point 2**
- Now, we calculate:
\[ I_2 = 1.3 \times \frac{100}{121} \approx 1.3 \times 0.8264 \approx 1.074 \, \text{kW/m}^2 \]

Thus, the average intensity of light at a distance of 22 cm from the lens is approximately **1.074 kW/m²**.

### Final Answer:
The average intensity of light at a distance of 22 cm from the lens is approximately **1.074 kW/m²**.