A concave mirror has a focal length 20 cm. The distance between the two positions of the object for which the image size is double of the object size is

Options

A. 20 cm
B. 40 cm
C. 30 cm
D. 60 cm

Answer

To solve the problem, we need to find the distance between the two positions of the object for which the image size is double that of the object size when using a concave mirror with a focal length of 20 cm.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Focal length (f) of the concave mirror = -20 cm (negative because it is a concave mirror).

2. **Understanding the Magnification Condition:**
- The image size is double the object size, which means the magnification (m) is -2 (the negative sign indicates that the image is real and inverted).
- Magnification (m) is given by the formula:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
where \( h' \) is the height of the image, \( h \) is the height of the object, \( v \) is the image distance, and \( u \) is the object distance.

3. **Set Up the Equation for Real Image:**
- For a real image where magnification is -2:
\[ -2 = -\frac{v}{u_1} \implies v = 2u_1 \]

4. **Apply the Mirror Formula:**
- The mirror formula is:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
- Substituting \( v = 2u_1 \) and \( f = -20 \):
\[ \frac{1}{-20} = \frac{1}{2u_1} + \frac{1}{u_1} \]
- This simplifies to:
\[ \frac{1}{-20} = \frac{1 + 2}{2u_1} = \frac{3}{2u_1} \]
- Cross-multiplying gives:
\[ 3 \cdot (-20) = 2u_1 \implies -60 = 2u_1 \implies u_1 = -30 \text{ cm} \]

5. **Set Up the Equation for Virtual Image:**
- For a virtual image where magnification is -2:
\[ -2 = -\frac{v}{u_2} \implies v = 2u_2 \]
- Using the mirror formula again:
\[ \frac{1}{-20} = \frac{1}{2u_2} + \frac{1}{u_2} \]
- This simplifies to:
\[ \frac{1}{-20} = \frac{3}{2u_2} \]
- Cross-multiplying gives:
\[ 3 \cdot (-20) = 2u_2 \implies -60 = 2u_2 \implies u_2 = -30 \text{ cm} \]

6. **Calculate the Distance Between the Two Object Positions:**
- The two object distances are \( u_1 = -30 \) cm and \( u_2 = -10 \) cm.
- The distance between the two positions is:
\[ |u_1 - u_2| = |-30 - (-10)| = |-30 + 10| = |-20| = 20 \text{ cm} \]

### Final Answer:
The distance between the two positions of the object for which the image size is double that of the object size is **20 cm**.

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