A plane mirror of circular shape with radius `r=20cm` is fixed to the ceiling .A bulb is to be placed on the axis of the mirror.A circular area of radius `R=1m` on the floor is to be illuminated after reflection of light from the mirror. The height of the room is `3m` What is maximum distance from the center of the mirror and the bulb so that the required area is illuminated?

Options

A. 75 cm
B. 70 cm
C. 80 cm
D. 100 cm

Answer

To solve the problem, we need to determine the maximum distance \( X \) from the center of the circular mirror to the bulb so that the light reflected from the mirror can illuminate a circular area of radius \( R = 1 \, m \) on the floor.

### Step 1: Understand the Geometry
- The circular mirror has a radius \( r = 20 \, cm = 0.2 \, m \).
- The height of the room is \( H = 3 \, m \).
- The illuminated area on the floor has a radius \( R = 1 \, m \).

### Step 2: Set Up the Problem
- The bulb is placed at a height \( H - x \) from the floor, where \( x \) is the distance from the center of the mirror to the bulb along the axis.
- The distance from the bulb to the edge of the illuminated area on the floor is \( R - r = 1 - 0.2 = 0.8 \, m \).

### Step 3: Use Similar Triangles
- We can use similar triangles to relate the height and the horizontal distances.
- For the larger triangle formed by the height of the room and the radius of the illuminated area:
\[ \tan(\theta) = \frac{0.8}{3} \]
- For the smaller triangle formed by the height from the bulb to the mirror and the radius of the mirror:
\[ \tan(\theta) = \frac{0.2}{x} \]

### Step 4: Set Up the Equations
From the two triangles, we can write:
\[ \frac{0.8}{3} = \frac{0.2}{x} \]

### Step 5: Cross Multiply and Solve for \( x \)
Cross-multiplying gives:
\[ 0.8x = 0.2 \cdot 3 \]
\[ 0.8x = 0.6 \]
\[ x = \frac{0.6}{0.8} = \frac{3}{4} \, m \]

### Conclusion
The maximum distance from the center of the mirror to the bulb is:
\[ x = 0.75 \, m \text{ or } 75 \, cm \]