A `4.5 cm` needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Options

A. 6 cm , 0.718
B. 6.7 cm , 0.558
C. Cannot be determined
D. NONE OF THESE

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Height of the needle (object height, H1) = 4.5 cm
- Object distance (U) = -12 cm (the negative sign indicates that the object is in front of the mirror)
- Focal length (F) = +15 cm (positive for a convex mirror)

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \( f \) = focal length
- \( v \) = image distance
- \( u \) = object distance

Substituting the values into the formula:
\[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-12} \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{12} \]

### Step 4: Finding a common denominator
The common denominator for 15 and 12 is 60. Thus:
\[ \frac{1}{15} = \frac{4}{60}, \quad \frac{1}{12} = \frac{5}{60} \]
So:
\[ \frac{1}{v} = \frac{4}{60} + \frac{5}{60} = \frac{9}{60} \]

### Step 5: Calculate the image distance (v)
Taking the reciprocal gives:
\[ v = \frac{60}{9} = \frac{20}{3} \approx 6.67 \text{ cm} \]

### Step 6: Calculate the magnification (m)
The magnification (m) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{20/3}{-12} = \frac{20}{3 \times 12} = \frac{20}{36} = \frac{5}{9} \approx 0.56 \]

### Step 7: Calculate the height of the image (H2)
The height of the image (H2) can be calculated using the magnification:
\[ H2 = m \times H1 = 0.56 \times 4.5 \approx 2.52 \text{ cm} \]

### Step 8: Describe what happens as the needle is moved farther from the mirror
As the needle (object) is moved farther from the convex mirror, the image distance (v) will increase, but the image will remain virtual, upright, and smaller than the object. The height of the image will decrease as the object distance increases.

### Summary of Results:
- Location of the image (v) = 6.67 cm (virtual)
- Magnification (m) = 0.56
- Height of the image (H2) = 2.52 cm