A concave mirror has a focal length of 20 cm. Find the position or positions of an object for which the image-size is double of the object-size.

Explanation

To solve the problem of finding the position or positions of an object for which the image size is double the object size in a concave mirror with a focal length of 20 cm, we will follow these steps:

### Step 1: Understand the given data
- Focal length (f) of the concave mirror = -20 cm (negative because it is a concave mirror).
- Magnification (m) = 2 (the image size is double the object size).

### Step 2: Set up the magnification formula
The magnification (m) for mirrors is given by the formula:
\[ m = -\frac{v}{u} \]
where:
- \( v \) = image distance
- \( u \) = object distance

Since the image size is double the object size, we can write:
\[ 2 = -\frac{v}{u} \]
This implies:
\[ v = -2u \]

### Step 3: Apply the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Substituting the values we have:
\[ \frac{1}{-20} = \frac{1}{-2u} + \frac{1}{u} \]

### Step 4: Solve for object distance (u)
Now, we need to simplify the equation:
1. Rewrite the right side:
\[ \frac{1}{-2u} + \frac{1}{u} = -\frac{1}{2u} + \frac{2}{2u} = \frac{1}{2u} \]

2. Substitute back into the mirror formula:
\[ \frac{1}{-20} = \frac{1}{2u} \]

3. Cross-multiply to solve for \( u \):
\[ 2u = -20 \]
\[ u = -10 \, \text{cm} \]

### Step 5: Consider the case where the image is real
Now, let’s consider the case where the image is real. We already have the magnification equation:
\[ v = -2u \]

Using the mirror formula again:
\[ \frac{1}{-20} = \frac{1}{-2u} + \frac{1}{u} \]

Substituting \( v = -2u \) into the mirror formula:
1. Rewrite the equation:
\[ \frac{1}{-20} = -\frac{1}{2u} + \frac{1}{u} \]

2. Combine the fractions:
\[ -\frac{1}{2u} + \frac{2}{2u} = \frac{1}{2u} \]

3. Substitute back:
\[ \frac{1}{-20} = \frac{1}{2u} \]

4. Cross-multiply:
\[ 2u = -20 \]
\[ u = -10 \, \text{cm} \]

### Step 6: Consider the case where the image is virtual
For the virtual image, we have:
\[ m = \frac{v}{u} \]
This means:
\[ 2 = \frac{v}{u} \]
Thus:
\[ v = 2u \]

Using the mirror formula:
\[ \frac{1}{-20} = \frac{1}{2u} + \frac{1}{u} \]

1. Combine the fractions:
\[ \frac{1}{2u} + \frac{2}{2u} = \frac{3}{2u} \]

2. Substitute back:
\[ \frac{1}{-20} = \frac{3}{2u} \]

3. Cross-multiply:
\[ 2u = -60 \]
\[ u = -30 \, \text{cm} \]

### Final Answer
The positions of the object for which the image size is double the object size are:
1. \( u = -10 \, \text{cm} \) (virtual image)
2. \( u = -30 \, \text{cm} \) (real image)