A spherical mirror forms a real image three times the linear size of the object. If the distance between the object and the image is 80 cm, the focal length of the mirror is

Options

A. 15 cm
B. `-15` cm
C. `30` cm
D. 40 cm

Explanation

To solve the problem, we need to find the focal length of a concave mirror that forms a real image three times the linear size of the object, with the distance between the object and image being 80 cm.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The image is real and three times the size of the object. Therefore, the magnification \( m = -\frac{h_i}{h_o} = -3 \) (since the image is inverted).
- The distance between the object and the image is given as 80 cm.

2. **Set Up the Relationships**:
- Let the distance of the object from the mirror be \( u \) and the distance of the image from the mirror be \( v \).
- From the magnification formula, we have:
\[ m = -\frac{v}{u} = -3 \implies v = -3u \]

3. **Use the Given Distance Between Object and Image**:
- The distance between the object and the image is given as:
\[ |v - u| = 80 \text{ cm} \]
- Substituting \( v = -3u \):
\[ |-3u - u| = 80 \]
\[ |-4u| = 80 \implies 4u = 80 \implies u = 20 \text{ cm} \]
- Therefore, substituting back to find \( v \):
\[ v = -3u = -3 \times 20 = -60 \text{ cm} \]

4. **Use the Mirror Formula**:
- The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
- Substitute \( u = -20 \) cm and \( v = -60 \) cm into the formula:
\[ \frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} \]
- Finding a common denominator (which is 60):
\[ \frac{1}{f} = -\frac{1}{60} - \frac{3}{60} = -\frac{4}{60} = -\frac{1}{15} \]

5. **Calculate the Focal Length**:
- Therefore, the focal length \( f \) is:
\[ f = -15 \text{ cm} \]
- The negative sign indicates that it is a concave mirror.

### Final Answer:
The focal length of the mirror is \( 15 \text{ cm} \) (concave mirror).

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