A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the image to be real, the focal length should be

Options

A. 5 cm
B. 15 cm
C. 20 cm
D. 30 cm

Answer

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

### Step 1: Understand the given information
- The object distance (u) is given as 20 cm. Since the object is in front of the mirror, we take it as negative:
\[ u = -20 \, \text{cm} \]
- The magnification (m) for a real image is given as 3 times the size of the object. For a real image, magnification is negative:
\[ m = -3 \]

### Step 2: Use the magnification formula
The magnification (m) is also defined as:
\[ m = \frac{V}{U} \]
Where:
- V is the image distance
- U is the object distance

Substituting the values we have:
\[ -3 = \frac{V}{-20} \]

### Step 3: Solve for V
From the equation above, we can rearrange it to find V:
\[ V = -3 \times (-20) \]
\[ V = 60 \, \text{cm} \]

### Step 4: Use the mirror formula
The mirror formula relates the object distance (U), image distance (V), and focal length (F):
\[ \frac{1}{F} = \frac{1}{V} + \frac{1}{U} \]

Substituting the values of V and U:
\[ \frac{1}{F} = \frac{1}{60} + \frac{1}{-20} \]

### Step 5: Simplify the equation
Convert \(\frac{1}{-20}\) to a common denominator with \(\frac{1}{60}\):
\[ \frac{1}{-20} = -\frac{3}{60} \]
So, we have:
\[ \frac{1}{F} = \frac{1}{60} - \frac{3}{60} \]
\[ \frac{1}{F} = -\frac{2}{60} \]
\[ \frac{1}{F} = -\frac{1}{30} \]

### Step 6: Find the focal length
Taking the reciprocal gives us the focal length:
\[ F = -30 \, \text{cm} \]

### Conclusion
The focal length of the concave mirror for the image to be real and three times the size of the object is:
\[ F = -30 \, \text{cm} \]