The image of an extended object placed in front of a concave mirror is formed at a distance of 40 cm from the object. If the image is 3 times bigger than the object, the magnitude of focal length of the mirror is

Options

A. 15 cm
B. 10 cm
C. 20 cm
D. 5 cm

Explanation

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

### Step 1: Understand the given information
- The distance between the object and the image is 40 cm.
- The image is 3 times larger than the object.

### Step 2: Define variables
Let:
- \( U \) = distance of the object from the mirror (in cm)
- \( V \) = distance of the image from the mirror (in cm)
- \( F \) = focal length of the mirror (in cm)

### Step 3: Use the magnification formula
The magnification \( m \) is given by:
\[ m = -\frac{V}{U} \]
Since the image is 3 times larger than the object:
\[ 3 = -\frac{V}{U} \implies V = -3U \]

### Step 4: Use the distance relationship
We know that the distance between the object and the image is 40 cm:
\[ |V - U| = 40 \]
Substituting \( V = -3U \):
\[ |-3U - U| = 40 \]
This simplifies to:
\[ |-4U| = 40 \implies 4U = 40 \implies U = -10 \text{ cm} \]
(We take the negative sign because the object is in front of the mirror.)

### Step 5: Calculate \( V \)
Now substituting \( U \) back into the equation for \( V \):
\[ V = -3(-10) = 30 \text{ cm} \]

### Step 6: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{F} = \frac{1}{V} + \frac{1}{U} \]
Substituting the values of \( U \) and \( V \):
\[ \frac{1}{F} = \frac{1}{30} + \frac{1}{-10} \]
Calculating the right side:
\[ \frac{1}{F} = \frac{1}{30} - \frac{3}{30} = \frac{-2}{30} = -\frac{1}{15} \]

### Step 7: Calculate \( F \)
Taking the reciprocal:
\[ F = -15 \text{ cm} \]
The magnitude of the focal length is:
\[ |F| = 15 \text{ cm} \]

### Final Answer
The magnitude of the focal length of the concave mirror is **15 cm**.

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