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 ?

Options

A. `-9.6 cm`
B. `-3.6 cm`
C. `-6.3 cm`
D. `-8.3 cm`

Explanation

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

### Step 1: Identify the given values
- Height of the object (h1) = 2 cm
- Height of the image (h2) = -3 cm (negative because the image is real)
- Object distance (u) = -16 cm (negative as per the sign convention for concave mirrors)

### Step 2: Use the magnification formula
The magnification (m) is given by the formula:
\[ m = \frac{h_2}{h_1} = -\frac{v}{u} \]

Substituting the known values:
\[ m = \frac{-3}{2} \]

### Step 3: Relate magnification to image distance (v)
From the magnification formula, we can express v in terms of u:
\[ -\frac{v}{u} = \frac{-3}{2} \]
\[ v = \frac{3}{2} \cdot u \]

Substituting u = -16 cm:
\[ v = \frac{3}{2} \cdot (-16) \]
\[ v = -24 \, \text{cm} \]

### Step 4: Calculate 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 v and u:
\[ \frac{1}{f} = \frac{1}{-24} + \frac{1}{-16} \]

Finding a common denominator (48):
\[ \frac{1}{f} = \frac{-2}{48} + \frac{-3}{48} \]
\[ \frac{1}{f} = \frac{-5}{48} \]

### Step 5: Solve for focal length (f)
Taking the reciprocal:
\[ f = \frac{-48}{5} \]
\[ f = -9.6 \, \text{cm} \]

### Step 6: Summary of results
- The focal length of the concave mirror is **-9.6 cm**.
- The position of the image is **-24 cm** (indicating that the image is formed on the same side as the object).