A concave mirror forms a virtual imageof size twice that of the object placed at a distance 5 cm from it. Find: a. the focal length of the mirror b. position of image.

Answer

To solve the problem step by step, we will use the mirror formula and the magnification formula.

### Step 1: Identify the given values
- The object distance (U) is given as 5 cm. Since the object is in front of the mirror, we take it as negative:
\[ U = -5 \, \text{cm} \]
- The magnification (HI/HO) is given as 2, meaning the image height (HI) is twice the object height (HO):
\[ \frac{HI}{HO} = 2 \]

### Step 2: Use the magnification formula
The magnification (m) is given by the formula:
\[ m = -\frac{V}{U} \]
Substituting the values we have:
\[ 2 = -\frac{V}{-5} \]
This simplifies to:
\[ 2 = \frac{V}{5} \]
Now, we can solve for V:
\[ V = 2 \times 5 = 10 \, \text{cm} \]

### Step 3: Find the focal length using the mirror formula
The mirror formula is given by:
\[ \frac{1}{F} = \frac{1}{V} + \frac{1}{U} \]
Substituting the values of U and V:
\[ \frac{1}{F} = \frac{1}{10} + \frac{1}{-5} \]
Calculating the right side:
\[ \frac{1}{F} = \frac{1}{10} - \frac{2}{10} = -\frac{1}{10} \]
Now, taking the reciprocal to find F:
\[ F = -10 \, \text{cm} \]

### Final Answers
- a. The focal length of the mirror (F) is \(-10 \, \text{cm}\).
- b. The position of the image (V) is \(10 \, \text{cm}\).