An object of size `10 cm` is placed at a distance of `50 cm` from a concave mirror of focal length `15 cm`. Calculate location, size and nature of the image.

Answer

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

### Step 1: Identify the given values
- Object size (h) = 10 cm
- Object distance (u) = -50 cm (the negative sign indicates that the object is in front of the mirror)
- Focal length (f) = -15 cm (the negative sign indicates that it is a concave mirror)

### 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 into the formula:
\[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-50} \]

### Step 3: Rearranging the equation
Rearranging the equation to solve for \(\frac{1}{v}\):
\[ \frac{1}{v} = \frac{1}{-15} + \frac{1}{50} \]

### Step 4: Finding a common denominator
To add the fractions, we need a common denominator. The least common multiple of 15 and 50 is 150. Thus, we rewrite the fractions:
\[ \frac{1}{-15} = \frac{-10}{150} \quad \text{and} \quad \frac{1}{50} = \frac{3}{150} \]
Now substituting back:
\[ \frac{1}{v} = \frac{-10 + 3}{150} = \frac{-7}{150} \]

### Step 5: Solve for v
Taking the reciprocal gives us:
\[ v = \frac{-150}{7} \approx -21.43 \text{ cm} \]
Thus, the image location is approximately \( -21.4 \text{ cm} \) from the mirror.

### Step 6: Calculate the magnification
The magnification (m) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values we have:
\[ m = -\frac{-21.43}{-50} = \frac{21.43}{50} \approx 0.4286 \]

### Step 7: Calculate the size of the image
The size of the image (h') can be calculated using:
\[ h' = m \times h \]
Substituting the values:
\[ h' = 0.4286 \times 10 \approx 4.286 \text{ cm} \]

### Step 8: Determine the nature of the image
Since the image distance (v) is negative, it indicates that the image is formed on the same side as the object, meaning it is a real image. Additionally, since the magnification is positive, the image is inverted.

### Summary of Results
- **Location of the image**: -21.4 cm (in front of the mirror)
- **Size of the image**: 4.29 cm
- **Nature of the image**: Real and inverted