An object is placed in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate two possible distances of the object from the mirror?

Options

A. `+40//5cm, -80//3cm`
B. `-40//3cm, -80//3cm`
C. `-20//3cm, -50//3cm`
D. `-40//3cm, +70//3cm`

Explanation

To solve the problem step by step, we need to find the object distance (U) for a concave mirror given its focal length and the magnification of the image formed.

### Step 1: Understand the given information
- Focal length (F) of the concave mirror = -20 cm (negative because it is a concave mirror)
- Magnification (m) = 3 (for a virtual image) or m = -3 (for a real image)

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

### Step 3: Case 1 - Virtual Image (m = 3)
For a virtual image, we have:
\[ m = 3 \]
Thus,
\[ 3 = -\frac{V}{U} \]
From this, we can express V in terms of U:
\[ V = -3U \]

### Step 4: Apply the mirror formula
The mirror formula is:
\[ \frac{1}{V} + \frac{1}{U} = \frac{1}{F} \]
Substituting V = -3U into the mirror formula:
\[ \frac{1}{-3U} + \frac{1}{U} = \frac{1}{-20} \]

### Step 5: Solve for U
To simplify, we find a common denominator:
\[ -\frac{1}{3U} + \frac{3}{3U} = -\frac{1}{20} \]
This simplifies to:
\[ \frac{2}{3U} = -\frac{1}{20} \]
Cross-multiplying gives:
\[ 2 \cdot 20 = -3U \]
\[ 40 = -3U \]
Thus,
\[ U = -\frac{40}{3} \text{ cm} \]

### Step 6: Case 2 - Real Image (m = -3)
For a real image, we have:
\[ m = -3 \]
Thus,
\[ -3 = -\frac{V}{U} \]
From this, we can express V in terms of U:
\[ V = 3U \]

### Step 7: Apply the mirror formula again
Using the mirror formula:
\[ \frac{1}{V} + \frac{1}{U} = \frac{1}{F} \]
Substituting V = 3U into the mirror formula:
\[ \frac{1}{3U} + \frac{1}{U} = \frac{1}{-20} \]

### Step 8: Solve for U again
Finding a common denominator:
\[ \frac{1}{3U} + \frac{3}{3U} = -\frac{1}{20} \]
This simplifies to:
\[ \frac{4}{3U} = -\frac{1}{20} \]
Cross-multiplying gives:
\[ 4 \cdot 20 = -3U \]
\[ 80 = -3U \]
Thus,
\[ U = -\frac{80}{3} \text{ cm} \]

### Final Results
The two possible distances of the object from the mirror are:
1. \( U = -\frac{40}{3} \text{ cm} \) (for a virtual image)
2. \( U = -\frac{80}{3} \text{ cm} \) (for a real image)