A convex mirror used for rear view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and magnification of the image.

Explanation

To solve the problem step by step, we will use the mirror formula and the magnification formula for a convex mirror.

### Step 1: Understand the given data
- Radius of curvature (R) = +3.00 m (positive for convex mirrors)
- Distance of the object (bus) from the mirror (U) = -5.00 m (negative as per the sign convention for mirrors)

### Step 2: Calculate the focal length (F)
The focal length (F) of a convex mirror is given by the formula:
\[ F = \frac{R}{2} \]
Substituting the value of R:
\[ F = \frac{3.00 \, \text{m}}{2} = 1.50 \, \text{m} \]

### Step 3: Use the mirror formula
The mirror formula is:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values of F and U:
\[ \frac{1}{1.50} = \frac{1}{v} + \frac{1}{-5} \]
This simplifies to:
\[ \frac{1}{v} = \frac{1}{1.50} + \frac{1}{5} \]

### Step 4: Calculate the right-hand side
First, find a common denominator to add the fractions:
- The least common multiple of 1.50 and 5 is 7.5.
- Convert the fractions:
\[ \frac{1}{1.50} = \frac{5}{7.5} \]
\[ \frac{1}{5} = \frac{1.5}{7.5} \]
Now, add them:
\[ \frac{1}{v} = \frac{5 + 1.5}{7.5} = \frac{6.5}{7.5} \]

### Step 5: Solve for v
Now, take the reciprocal to find v:
\[ v = \frac{7.5}{6.5} \]
Calculating this gives:
\[ v \approx 1.15 \, \text{m} \]

### Step 6: Determine the nature of the image
Since v is positive, the image is virtual and located behind the mirror.

### Step 7: Calculate the magnification (m)
The magnification (m) is given by the formula:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{1.15}{-5} = \frac{1.15}{5} \]
Calculating this gives:
\[ m \approx 0.23 \]

### Step 8: Determine the nature of the image based on magnification
- Since the magnification is positive, the image is upright.
- Since the magnification is less than 1, the image is diminished.

### Final Summary
- Position of the image (v): 1.15 m (behind the mirror)
- Nature of the image: Virtual, upright, and diminished
- Magnification: 0.23

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