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

Explanation

To solve the problem, we need to find the distance between two positions of an object for which the image size is double the object size 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's a concave mirror).
- Magnification (M) = ±2 (since the image size is double the object size).

2. **Case 1: Virtual Image (M = +2)**
- For a virtual image, we use the magnification formula:
\[ M = \frac{v}{u} \]
where \(v\) is the image distance and \(u\) is the object distance. Rearranging gives:
\[ v = 2u \]
- Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
- Substitute \(f = -20\) and \(v = 2u\):
\[ \frac{1}{-20} = \frac{1}{2u} + \frac{1}{u} \]
- Finding a common denominator:
\[ \frac{1}{-20} = \frac{1 + 2}{2u} = \frac{3}{2u} \]
- Cross-multiplying gives:
\[ 2u = -60 \implies u = -30 \text{ cm} \]

3. **Case 2: Real Image (M = -2)**
- For a real image, we use the same magnification formula:
\[ M = \frac{v}{u} \implies v = -2u \]
- Again, using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
- Substitute \(f = -20\) and \(v = -2u\):
\[ \frac{1}{-20} = \frac{1}{-2u} + \frac{1}{u} \]
- Finding a common denominator:
\[ \frac{1}{-20} = \frac{-1 + 2}{2u} = \frac{1}{2u} \]
- Cross-multiplying gives:
\[ 2u = -20 \implies u = -10 \text{ cm} \]

4. **Calculate the Distance Between the Two Positions:**
- We have two object positions:
- \(u_1 = -30 \text{ cm}\) (for virtual image)
- \(u_2 = -10 \text{ cm}\) (for real image)
- The distance \(d\) between the two positions is:
\[ d = |u_1 - u_2| = |-30 - (-10)| = |-30 + 10| = |-20| = 20 \text{ cm} \]

### Final Answer:
The distance between the two positions of the object is **20 cm**.