There is a concave mirror of focal length f and an object of length 2.5 cm is placed at a distance 1.5 f from concave mirror. Length of the object is perpendicular to the principal axis. What is the length of image in cm ?
`|{:(0,1,2,3,4,5,6,7,8,9):}|`

Explanation

To find the length of the image formed by a concave mirror when an object is placed in front of it, we will follow these steps:

### Step 1: Identify the Given Values
- Focal length of the concave mirror, \( f \)
- Length of the object, \( h_o = 2.5 \) cm
- Object distance, \( u = -1.5f \) (negative as per the sign convention for concave mirrors)

### Step 2: Use the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \( f \) is the focal length
- \( v \) is the image distance
- \( u \) is the object distance

Substituting the values into the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{-1.5f} \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{1.5f} \]
To combine the fractions, we find a common denominator:
\[ \frac{1}{v} = \frac{1.5 + 1}{1.5f} = \frac{2.5}{1.5f} \]

### Step 4: Solve for Image Distance \( v \)
Taking the reciprocal to find \( v \):
\[ v = \frac{1.5f}{2.5} = \frac{3f}{5} \]
Since \( v \) is positive, the image is real and formed on the same side as the object.

### Step 5: Calculate the Magnification
The magnification \( m \) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values of \( v \) and \( u \):
\[ m = -\frac{\frac{3f}{5}}{-1.5f} = \frac{3f}{5 \times 1.5f} = \frac{3}{7.5} = \frac{2}{5} \]

### Step 6: Calculate the Height of the Image
The height of the image \( h_i \) can be calculated using the magnification:
\[ h_i = m \cdot h_o \]
Substituting the values:
\[ h_i = \frac{2}{5} \times 2.5 = 1 \text{ cm} \]

### Final Answer
The length of the image is \( 1 \) cm.