An object is placed at 10 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 = 30 cm, m=3 (real, inverted)
B. v = 20 cm, m = 3 (virtual, erect)
C. v = 10 cm, same size (real, inverted)
D. v= 10 cm, same size (virtual, erect)

Explanation

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

### Step 1: Identify the given values
- Object distance (u) = -10 cm (negative because the object is in front of the mirror)
- Radius of curvature (R) = 15 cm

### 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:
\[ f = -\frac{15}{2} = -7.5 \, \text{cm} \]

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

Substituting the values of f and u:
\[ \frac{1}{v} = \frac{1}{-7.5} - \frac{1}{-10} \]

Calculating the right-hand side:
\[ \frac{1}{v} = -\frac{1}{7.5} + \frac{1}{10} \]

Finding a common denominator (which is 30):
\[ \frac{1}{v} = -\frac{4}{30} + \frac{3}{30} = -\frac{1}{30} \]

Taking the reciprocal to find \( v \):
\[ v = -30 \, \text{cm} \]

### Step 4: Calculate the magnification (m)
The magnification (m) is given by the formula:
\[ m = \frac{v}{u} \]
Substituting the values of v and u:
\[ m = \frac{-30}{-10} = 3 \]

### Final Results
- Position of the image (v) = -30 cm (indicating that the image is on the same side as the object)
- Magnification (m) = 3 (indicating that the image is three times larger than the object)

### Summary
The position of the image is at -30 cm, and the magnification is 3.