Object is placed 15 cm from a concave mirror of focal length 10 cm, then the nature of the image formed will be

Options

A. magnified and inverted
B. magnified and erect
C. small in size and inverted
D. small in size and erect

Answer

To determine the nature of the image formed by a concave mirror when an object is placed 15 cm from it and the focal length is 10 cm, we can follow these steps:

### Step 1: Identify the given values
- Object distance (u) = -15 cm (negative because it is in front of the mirror)
- Focal length (f) = -10 cm (negative for concave mirrors)

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-15} \]

### Step 3: Rearranging the equation
Rearranging the equation to solve for \( \frac{1}{v} \):
\[ \frac{1}{v} = \frac{1}{-10} + \frac{1}{15} \]

### Step 4: Finding a common denominator
To add the fractions, we need a common denominator. The least common multiple of 10 and 15 is 30:
\[ \frac{1}{-10} = \frac{-3}{30}, \quad \frac{1}{15} = \frac{2}{30} \]
Thus,
\[ \frac{1}{v} = \frac{-3}{30} + \frac{2}{30} = \frac{-1}{30} \]

### Step 5: Solve for v
Now, we can find \( v \):
\[ v = -30 \text{ cm} \]
The negative sign indicates that the image is real and located on the same side as the object.

### Step 6: Determine the nature of the image
Since the image distance \( v \) is negative, the image is real. Additionally, we can find the magnification (m) using the formula:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{-30}{-15} = -2 \]
The negative sign indicates that the image is inverted, and the magnitude of 2 indicates that the image is magnified.

### Conclusion
The nature of the image formed is **real, inverted, and magnified**.

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