An object `2 cm` high is placed at a distance of `16 cm` from a concave mirror, which produces a real image `3 cm` high. What is thr focal length of the mirror ? Find the position of the image ?

Explanation

To find the focal length of the concave mirror and the position of the image, we can use the mirror formula and magnification formula. Let's solve this step by step.

### Step 1: Given Data
- Height of the object, \( h_o = 2 \, \text{cm} \)
- Distance of the object from the mirror, \( u = -16 \, \text{cm} \) (negative because the object is in front of the mirror)
- Height of the image, \( h_i = -3 \, \text{cm} \) (negative because the image is real and inverted)

### Step 2: Magnification Formula
The magnification \( m \) produced by a mirror is given by:
\[ m = \frac{h_i}{h_o} = \frac{-v}{u} \]

Substitute the given values:
\[ \frac{-3}{2} = \frac{-v}{-16} \]

### Step 3: Solve for Image Distance \( v \)
\[ \frac{-3}{2} = \frac{v}{16} \]
\[ v = 16 \times \frac{-3}{2} \]
\[ v = -24 \, \text{cm} \]

So, the image is located 24 cm in front of the mirror (on the same side as the object).

### Step 4: Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Substitute the values of \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{-24} + \frac{1}{-16} \]
\[ \frac{1}{f} = -\frac{1}{24} - \frac{1}{16} \]

### Step 5: Simplify the Equation
Find a common denominator for the fractions:
\[ \frac{1}{f} = -\frac{2}{48} - \frac{3}{48} \]
\[ \frac{1}{f} = -\frac{5}{48} \]

### Step 6: Solve for Focal Length \( f \)
\[ f = -\frac{48}{5} \]
\[ f = -9.6 \, \text{cm} \]

So, the focal length of the mirror is \( -9.6 \, \text{cm} \).

### Summary
- The position of the image is \( -24 \, \text{cm} \) (in front of the mirror).
- The focal length of the concave mirror is \( -9.6 \, \text{cm} \).