A concave mirror forms an erect image twice the size of a object. The object distance from the mirror is ______

Options

A. `f/2`
B. 2f
C. `(3f)/2`
D. `(2f)/3`

Answer

To solve the problem of finding the object distance from a concave mirror that forms an erect image twice the size of the object, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Magnification**:
- The magnification (m) of a mirror is defined as the ratio of the height of the image (SI) to the height of the object (SO).
- Given that the image is twice the size of the object, we can express this as:
\[ m = \frac{SI}{SO} = 2 \]

2. **Relating Magnification to Object and Image Distances**:
- Magnification can also be expressed in terms of object distance (U) and image distance (V):
\[ m = -\frac{V}{U} \]
- Since we know that the magnification is 2, we can set up the equation:
\[ 2 = -\frac{V}{U} \]
- Rearranging gives us:
\[ V = -2U \]

3. **Using the Mirror Formula**:
- The mirror formula relates the object distance (U), image distance (V), and focal length (F) of the mirror:
\[ \frac{1}{F} = \frac{1}{V} + \frac{1}{U} \]
- Substituting \(V = -2U\) into the mirror formula:
\[ \frac{1}{F} = \frac{1}{-2U} + \frac{1}{U} \]

4. **Finding a Common Denominator**:
- The common denominator for the right side is \(-2U\):
\[ \frac{1}{F} = \frac{1}{-2U} + \frac{2}{-2U} = \frac{-1 + 2}{-2U} = \frac{1}{-2U} \]

5. **Solving for U**:
- Now we can equate the two sides:
\[ \frac{1}{F} = \frac{1}{-2U} \]
- Cross-multiplying gives:
\[ -2U = F \]
- Therefore:
\[ U = -\frac{F}{2} \]

6. **Conclusion**:
- The object distance from the mirror is:
\[ U = -\frac{F}{2} \]
- Since the object distance is typically expressed as a positive value, we can state that the object distance is \( \frac{F}{2} \) (considering the magnitude).

### Final Answer:
The object distance from the mirror is \( \frac{F}{2} \).

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