An object is kept on the principal axis of a concave mirror of focal length 10cm, at a distance of 15 cm from its pole. The image formed by the mirror is:

Options

A. Virtual and magnified
B. Virtual and diminished
C. Real and magnified
D. Real and diminished

Answer

To solve the problem of finding the image formed by a concave mirror, we will use the mirror formula and the magnification formula. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Focal length of the concave mirror (f) = -10 cm (negative because it is a concave mirror)
- Object distance (u) = -15 cm (negative as per the sign convention for 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

### Step 3: Substitute the known values into the mirror formula
Substituting the values of \( f \) and \( u \):
\[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-15} \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{-10} + \frac{1}{15} \]

### Step 5: Find a common denominator and simplify
To add the fractions, find a common denominator (which is 30):
\[ \frac{1}{-10} = \frac{-3}{30}, \quad \frac{1}{15} = \frac{2}{30} \]
Now substituting these values:
\[ \frac{1}{v} = \frac{-3}{30} + \frac{2}{30} = \frac{-1}{30} \]

### Step 6: Solve for v
Taking the reciprocal gives:
\[ v = -30 \text{ cm} \]

### Step 7: Determine the nature of the image
Since \( v \) is negative, this indicates that the image is real and located on the same side as the object.

### Step 8: Calculate the magnification
The magnification (m) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{-30}{-15} = -2 \]

### Step 9: Interpret the magnification
Since the magnification is greater than 1 (in absolute value), the image is inverted and magnified.

### Conclusion
The image formed by the concave mirror is real, inverted, and magnified.

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