A person wants a real image of his own, 3 times enlarged. Where should he stand in front of a concave mirror of radius of curvature of 30cm.

Options

A. 90cm
B. 10cm
C. 20cm
D. 30cm

Explanation

To solve the problem of where a person should stand in front of a concave mirror to obtain a real image that is 3 times enlarged, we can follow these steps:

### Step 1: Understand the magnification
The magnification (m) for a real image is given as -3 (since real images formed by concave mirrors are inverted). Thus, we have:
\[ m = -\frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance.

### Step 2: Relate image distance and object distance
From the magnification formula, we can express \( v \) in terms of \( u \):
\[ -3 = -\frac{v}{u} \]
This simplifies to:
\[ v = 3u \]

### Step 3: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
For a concave mirror, the focal length \( f \) is negative. The radius of curvature \( R \) is given as 30 cm, so:
\[ f = -\frac{R}{2} = -\frac{30}{2} = -15 \text{ cm} \]

### Step 4: Substitute \( v \) into the mirror formula
Now we can substitute \( v = 3u \) into the mirror formula:
\[ \frac{1}{-15} = \frac{1}{3u} + \frac{1}{u} \]

### Step 5: Combine the fractions
Combining the fractions on the right side:
\[ \frac{1}{3u} + \frac{1}{u} = \frac{1 + 3}{3u} = \frac{4}{3u} \]
Thus, we have:
\[ \frac{1}{-15} = \frac{4}{3u} \]

### Step 6: Solve for \( u \)
Cross-multiplying gives:
\[ 3u = -15 \times 4 \]
\[ 3u = -60 \]
\[ u = -20 \text{ cm} \]

### Step 7: Interpret the result
The negative sign indicates that the object (the person) is located 20 cm in front of the mirror.

### Final Answer
The person should stand 20 cm in front of the concave mirror.

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