An object is placed at 4 cm distance in front of a concave mirror of radius of curvature 24 cm. Find the position of image. Is the image magnified?

Answer

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Identify the given data
- Object distance (u) = -4 cm (the negative sign indicates that the object is in front of the mirror)
- Radius of curvature (R) = 24 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:
\[ f = \frac{24 \, \text{cm}}{2} = 12 \, \text{cm} \]
Since it is a concave mirror, we take the focal length as negative:
\[ f = -12 \, \text{cm} \]

### Step 3: Use the mirror formula to find the image distance (v)
The mirror formula is:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values of f and u:
\[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-4} \]
This simplifies to:
\[ \frac{1}{-12} = \frac{1}{v} - \frac{1}{4} \]

### Step 4: Rearranging the equation to solve for v
To isolate \( \frac{1}{v} \), we can add \( \frac{1}{4} \) to both sides:
\[ \frac{1}{v} = \frac{1}{-12} + \frac{1}{4} \]
Finding a common denominator (which is 12):
\[ \frac{1}{v} = \frac{-1}{12} + \frac{3}{12} \]
\[ \frac{1}{v} = \frac{2}{12} = \frac{1}{6} \]

### Step 5: Solve for v
Taking the reciprocal gives:
\[ v = 6 \, \text{cm} \]

### Step 6: Determine the position of the image
Since \( v \) is positive, this indicates that the image is formed on the same side as the object, which means it is located 6 cm behind the mirror.

### Step 7: Calculate the magnification (M)
The magnification (M) is given by the formula:
\[ M = -\frac{v}{u} \]
Substituting the values of v and u:
\[ M = -\frac{6}{-4} = \frac{6}{4} = 1.5 \]

### Step 8: Conclusion about the magnification
Since the magnification is greater than 1 (M = 1.5), we conclude that the image is magnified.

### Final Answer
- The position of the image is at 6 cm behind the mirror.
- The image is magnified.

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