A concave mirror forms a real image of an object placed in front of it at a distance 30 cm, of size three times the size of object. Find a. the focal length of mirror b. position of image.

Answer

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

### Step 1: Identify the given values
- The object distance (u) is given as 30 cm. Since the object is in front of the mirror, we take it as negative:
\[ u = -30 \, \text{cm} \]
- The size of the image (h_i) is three times the size of the object (h_o):
\[ h_i = 3h_o \]

### Step 2: Use the magnification formula
The magnification (m) is given by:
\[ m = \frac{h_i}{h_o} = -\frac{V}{U} \]
Substituting the values we have:
\[ 3 = -\frac{V}{-30} \]
This simplifies to:
\[ 3 = \frac{V}{30} \]
Now, solving for V:
\[ V = 3 \times 30 = 90 \, \text{cm} \]

### Step 3: Use the mirror formula to find the focal length (f)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{V} + \frac{1}{U} \]
Substituting the values of V and U:
\[ \frac{1}{f} = \frac{1}{90} + \frac{1}{-30} \]
Finding a common denominator (which is 90):
\[ \frac{1}{f} = \frac{1}{90} - \frac{3}{90} = \frac{1 - 3}{90} = \frac{-2}{90} \]
Thus,
\[ f = \frac{90}{-2} = -45 \, \text{cm} \]

### Step 4: Summarize the findings
- The focal length of the concave mirror is:
\[ \text{Focal length} (f) = -45 \, \text{cm} \]
- The position of the image is:
\[ \text{Position of image} (V) = 90 \, \text{cm} \, \text{(in front of the mirror)} \]

### Final Answers
a. The focal length of the mirror is -45 cm.
b. The position of the image is 90 cm in front of the mirror.