In a car a rear view mirror having a radius of curvature 1.50 m forms a virtual image of a bus located 10.0 m from the mirror. The factor by which the mirror magnifies the size of the bus is close to

Options

A. 0.06
B. 0.07
C. 0.08
D. 0.09

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Radius of curvature (R) = 1.50 m
- Object distance (u) = -10.0 m (the negative sign indicates that the object is in front of the mirror)

### Step 2: Calculate the focal length (f) of the mirror
The focal length (f) of a mirror is given by the formula:
\[ f = \frac{R}{2} \]
Substituting the value of R:
\[ f = \frac{1.50}{2} = 0.75 \, \text{m} \]

### Step 3: Use the mirror formula to find the image distance (v)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging the formula to find v:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
Substituting the values of f and u:
\[ \frac{1}{v} = \frac{1}{0.75} - \frac{1}{-10} \]
\[ \frac{1}{v} = \frac{4}{3} + \frac{1}{10} \]

### Step 4: Find a common denominator and simplify
The common denominator for 3 and 10 is 30:
\[ \frac{1}{v} = \frac{40}{30} + \frac{3}{30} = \frac{43}{30} \]
Now, taking the reciprocal to find v:
\[ v = \frac{30}{43} \, \text{m} \]

### Step 5: Calculate the magnification (m)
The magnification (m) for mirrors is given by the formula:
\[ m = -\frac{v}{u} \]
Substituting the values of v and u:
\[ m = -\frac{\frac{30}{43}}{-10} \]
\[ m = \frac{30}{430} = \frac{3}{43} \]

### Step 6: Approximate the magnification
Calculating the approximate value:
\[ m \approx 0.07 \]

### Final Answer
The factor by which the mirror magnifies the size of the bus is approximately **0.07**.

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