A point source of light is kept at a distance of `15cm` from a converging lens,on its optical axis.The focal length of the lens is `10cm` and its diameter is `3cm` ,A screen is placed on the other side of the lens ,perpendicular to the axis of lens,at a distance `20cm` from it.Then find the area of the illuminated part of the screen?

Options

A. `pi/4cm^2`
B. `pi/6cm^2`
C. `pi/2cm^2`
D. `pi/3cm^2`

Answer

To find the area of the illuminated part of the screen, we will follow these steps:

### Step 1: Identify the given values
- Object distance (u) = -15 cm (negative as per the sign convention)
- Focal length (f) = 10 cm (positive for a converging lens)
- Distance of the screen from the lens = 20 cm

### Step 2: Use the lens formula to find the image distance (v)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

Substituting the values:
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{-15} \]
Finding a common denominator (30):
\[ \frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \]
Thus,
\[ v = 30 \text{ cm} \]

### Step 3: Analyze the position of the screen
The image distance (30 cm) is greater than the distance of the screen (20 cm). This means that the screen will not capture the full image formed by the lens.

### Step 4: Determine the diameter of the illuminated area on the screen using similar triangles
Using similar triangles, we can relate the diameter of the lens (D_lens = 3 cm) to the diameter of the illuminated area (D_screen) on the screen.

From the lens to the image:
- Distance from lens to image = 30 cm
- Distance from lens to screen = 20 cm

Using the property of similar triangles:
\[ \frac{D_{screen}}{20} = \frac{D_{lens}}{30} \]
Substituting the known diameter of the lens:
\[ \frac{D_{screen}}{20} = \frac{3}{30} \]
Cross-multiplying gives:
\[ D_{screen} = 20 \times \frac{3}{30} = 2 \text{ cm} \]

### Step 5: Calculate the area of the illuminated part of the screen
The area \( A \) of the illuminated part of the screen can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi D^2}{4} \]
Substituting \( D_{screen} = 2 \text{ cm} \):
\[ A = \frac{\pi (2)^2}{4} = \frac{\pi \times 4}{4} = \pi \text{ cm}^2 \]

### Final Answer
The area of the illuminated part of the screen is \( \pi \text{ cm}^2 \).

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