An object is placed on the principal axis of a concave mirror at a distance of 1.5 f(f is the focal length). The image will be at,

Options

A. 1.5f
B. `-1.5f`
C. 3f
D. `-3f`

Explanation

To solve the problem of finding the image distance when an object is placed at a distance of 1.5f from a concave mirror, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The object distance \( u = -1.5f \) (negative because the object is in front of the mirror).
- The focal length \( f \) of the concave mirror is also negative, so \( f = -f \).

2. **Use the Mirror Formula:**
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Here, we need to find the image distance \( v \).

3. **Substitute the Values into the Mirror Formula:**
Substitute \( u \) and \( f \) into the mirror formula:
\[ \frac{1}{-f} = \frac{1}{-1.5f} + \frac{1}{v} \]

4. **Rearranging the Equation:**
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{-f} + \frac{1}{1.5f} \]

5. **Finding a Common Denominator:**
The common denominator for the fractions on the right side is \( -1.5f \):
\[ \frac{1}{v} = \frac{-1.5 + 1}{-1.5f} \]
Simplifying the numerator:
\[ \frac{1}{v} = \frac{-0.5}{-1.5f} \]

6. **Simplifying Further:**
This simplifies to:
\[ \frac{1}{v} = \frac{0.5}{1.5f} \]
Which can be simplified to:
\[ \frac{1}{v} = \frac{1}{3f} \]

7. **Finding \( v \):**
Taking the reciprocal gives:
\[ v = 3f \]
Since we are dealing with a concave mirror, the image distance \( v \) will be negative:
\[ v = -3f \]

### Final Answer:
The image will be at a distance of \( -3f \) from the mirror.