For a concave mirror of focal length f, image is 2 times larger. Then the object distance from the mirror can be

Options

A. virtual image is always larger in size
B. real image is always smaller in size
C. real image is always larger in size
D. real image may be smaller or larger in size

Answer

To find the object distance from a concave mirror when the image is 2 times larger than the object, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Magnification**:
The magnification (m) is given as 2. For mirrors, magnification is defined as:
\[ m = -\frac{v}{u} \]
where \(v\) is the image distance and \(u\) is the object distance. Since the image is 2 times larger, we can write:
\[ m = \pm 2 \]
Therefore, we have:
\[ -\frac{v}{u} = \pm 2 \]

2. **Set Up the Two Cases**:
We can consider two cases based on the magnification:
- Case 1: \(m = 2\) (real image)
- Case 2: \(m = -2\) (virtual image)

3. **Case 1: Real Image (m = 2)**:
From the magnification formula:
\[ -\frac{v}{u} = 2 \implies v = -2u \]
Now, we use the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting \(v = -2u\):
\[ \frac{1}{f} = \frac{1}{-2u} + \frac{1}{u} \]
Simplifying this:
\[ \frac{1}{f} = -\frac{1}{2u} + \frac{2}{2u} = \frac{1}{2u} \]
Therefore:
\[ f = 2u \implies u = \frac{f}{2} \]

4. **Case 2: Virtual Image (m = -2)**:
From the magnification formula:
\[ -\frac{v}{u} = -2 \implies v = 2u \]
Again, using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting \(v = 2u\):
\[ \frac{1}{f} = \frac{1}{2u} + \frac{1}{u} \]
Simplifying this:
\[ \frac{1}{f} = \frac{1}{2u} + \frac{2}{2u} = \frac{3}{2u} \]
Therefore:
\[ f = \frac{2u}{3} \implies u = \frac{3f}{2} \]

5. **Final Results**:
The object distance can be:
- \(u = \frac{f}{2}\) (for the real image)
- \(u = \frac{3f}{2}\) (for the virtual image)

### Summary:
The object distance from the mirror can be either \( \frac{f}{2} \) or \( \frac{3f}{2} \).