An obj ect is placed at 5 cm infront of a concave mirror of radius of curvature 15 cm. The position of image (v) and its magnification (m) are

Options

A. v = 15 cm, m = 3 (virtual, erect)
B. ) v = 5 cm, same size (virtual, erect)
C. v = 5 cm, same size (real, inverted)
D. v = 15 cm, m = 3 (real, inverted)

Explanation

To solve the problem step by step, we will follow the mirror formula and the magnification formula for a concave mirror.

### Step 1: Identify the given values
- Object distance (u) = -5 cm (the negative sign indicates that the object is in front of the mirror)
- Radius of curvature (R) = -15 cm (the negative sign indicates that it is a concave mirror)

### Step 2: Calculate the focal length (f)
The focal length (f) of a concave mirror is given by the formula:
\[ f = \frac{R}{2} \]
Substituting the value of R:
\[ f = \frac{-15 \, \text{cm}}{2} = -7.5 \, \text{cm} \]

### Step 3: Use the mirror formula to find the image distance (v)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-7.5} = \frac{1}{v} + \frac{1}{-5} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-7.5} + \frac{1}{5} \]

### Step 5: Finding a common denominator
The common denominator between -7.5 and 5 is 15. Thus, we rewrite the fractions:
\[ \frac{1}{-7.5} = \frac{-2}{15} \quad \text{and} \quad \frac{1}{5} = \frac{3}{15} \]

### Step 6: Substitute back into the equation
Now substituting these values back:
\[ \frac{1}{v} = \frac{-2}{15} + \frac{3}{15} = \frac{1}{15} \]

### Step 7: Solve for v
Taking the reciprocal gives:
\[ v = 15 \, \text{cm} \]

### Step 8: Calculate the magnification (m)
The magnification (m) is given by the formula:
\[ m = -\frac{v}{u} \]
Substituting the values of v and u:
\[ m = -\frac{15}{-5} = 3 \]

### Final Results
- The position of the image (v) = 15 cm
- The magnification (m) = 3