A point object is kept at (1,0,0). A circular plane mirror of radius 1m is kept in yz-plane such that its centre is at the origin. The reflecting side faces positive x-axis At which of the following points can the image of the object be seen?

Options

A. (-0.5,0,0.5)
B. (2,2,2)
C. (1,1.5,1.5)
D. (1,-1,1.5)

Explanation

To solve the problem of determining at which points the image of the object can be seen, we follow these steps:

### Step 1: Understand the setup
- The point object is located at (1, 0, 0).
- The circular plane mirror is in the yz-plane, centered at the origin (0, 0, 0), with a radius of 1 meter.
- The reflecting side of the mirror faces the positive x-axis.

### Step 2: Determine the image position
- The image of the object in a plane mirror is located at an equal distance behind the mirror. Since the object is at (1, 0, 0), the image will be at (-1, 0, 0).

### Step 3: Determine the field of view
- The field of view from the mirror can be visualized as a cone with its vertex at the image point (-1, 0, 0) and opening towards the positive x-axis.
- The radius of the mirror is 1 meter, and the distance from the mirror to the object is 1 meter. Therefore, the cone will have a height of 1 meter and a base radius of 1 meter.

### Step 4: Calculate the cone's dimensions
- The cone's angle can be determined using the relationship of the radius and height. The angle θ can be calculated using:
\[ \tan(\theta) = \frac{\text{radius}}{\text{height}} = \frac{1}{1} = 1 \implies \theta = 45^\circ \]
- The cone will extend from the image point (-1, 0, 0) and can be described mathematically.

### Step 5: Identify the points given in the options
We will analyze each option to see if it lies within the field of view of the cone.

1. **Option A: (-0.5, 0, 0.5)**
- This point is behind the mirror (x < 0), hence it cannot see the image.

2. **Option B: (2, 2, 2)**
- This point is at x = 2. The distance from the image (-1, 0, 0) to this point can be calculated:
\[ \text{Distance} = \sqrt{(2 - (-1))^2 + (2 - 0)^2 + (2 - 0)^2} = \sqrt{(3)^2 + (2)^2 + (2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17} \]
- The radius of the cone at x = 2 is 3 (since the cone opens outwards), so this point can see the image.

3. **Option C: (1, 1.5, 1.5)**
- This point is at x = 1. The distance from the image (-1, 0, 0) to this point is:
\[ \text{Distance} = \sqrt{(1 - (-1))^2 + (1.5 - 0)^2 + (1.5 - 0)^2} = \sqrt{(2)^2 + (1.5)^2 + (1.5)^2} = \sqrt{4 + 2.25 + 2.25} = \sqrt{8.5} \]
- The radius of the cone at x = 1 is 2, which is less than the distance calculated. Thus, this point cannot see the image.

4. **Option D: (1, -1, 1.5)**
- This point is also at x = 1. The distance from the image (-1, 0, 0) is:
\[ \text{Distance} = \sqrt{(1 - (-1))^2 + (-1 - 0)^2 + (1.5 - 0)^2} = \sqrt{(2)^2 + (-1)^2 + (1.5)^2} = \sqrt{4 + 1 + 2.25} = \sqrt{7.25} \]
- The radius of the cone at x = 1 is still 2, which is greater than the distance calculated. Thus, this point can see the image.

### Conclusion
The points where the image can be seen are:
- **Option B: (2, 2, 2)**
- **Option D: (1, -1, 1.5)**