A concave mirror produces a real image 10 mm tall, of an object 2.5 mm tall placed at 5 cm from the mirror. Calculate focal length of the mirror and the position of the image?

Options

A. `-4cm ; -20cm`
B. `-40cm ; -20cm`
C. `-2cm ; -10cm`
D. `-5cm ; -15cm`

Explanation

Let's solve the problem step by step:

### Given:
- Height of the object (\( h_o \)) = 2.5 mm
- Height of the image (\( h_i \)) = -10 mm (since the image is real and inverted)
- Object distance (\( u \)) = -5 cm (object distance is taken as negative in mirror formula)

### To Find:
1. Focal length of the mirror (\( f \))
2. Position of the image (\( v \))

### Step 1: Calculate the Magnification (\( m \))
The magnification (\( m \)) is given by the ratio of the height of the image to the height of the object:
\[ m = \frac{h_i}{h_o} \]

Substitute the given values:
\[ m = \frac{-10 \text{ mm}}{2.5 \text{ mm}} \]
\[ m = -4 \]

### Step 2: Relate Magnification to Object and Image Distances
Magnification (\( m \)) is also given by:
\[ m = \frac{-v}{u} \]

We already have \( m = -4 \) and \( u = -5 \) cm. Substitute these values into the equation:
\[ -4 = \frac{-v}{-5} \]
\[ -4 = \frac{v}{5} \]

Solve for \( v \):
\[ v = -4 \times 5 \]
\[ v = -20 \text{ cm} \]

So, the position of the image is \( -20 \) cm.

### Step 3: Use the Mirror Formula to Find the Focal Length
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Substitute the values of \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{-20} + \frac{1}{-5} \]

Convert to a common denominator:
\[ \frac{1}{f} = \frac{-1}{20} + \frac{-4}{20} \]
\[ \frac{1}{f} = \frac{-1 - 4}{20} \]
\[ \frac{1}{f} = \frac{-5}{20} \]
\[ \frac{1}{f} = \frac{-1}{4} \]

Solve for \( f \):
\[ f = -4 \text{ cm} \]

### Final Answers:
1. The focal length of the mirror (\( f \)) is \( -4 \) cm.
2. The position of the image (\( v \)) is \( -20 \) cm.