The focal length of a concave mirror is 12 cm . Where should an object of length 4 cm be placed so that a real image of 1 cm length is formed ?

Options

A. 48 cm
B. 3 cm
C. 60 cm
D. 15 cm

Answer

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

### Step 1: Identify the given values
- Focal length of the concave mirror, \( f = -12 \, \text{cm} \) (negative because it's a concave mirror).
- Height of the object, \( h_o = 4 \, \text{cm} \).
- Height of the image, \( h_i = -1 \, \text{cm} \) (negative because the image is real and inverted).

### Step 2: Calculate the magnification
The magnification \( M \) can be calculated using the formula:
\[ M = \frac{h_i}{h_o} = \frac{-1}{4} = -\frac{1}{4} \]
We also know that magnification can be expressed in terms of image distance \( V \) and object distance \( U \):
\[ M = -\frac{V}{U} \]
Setting the two expressions for magnification equal gives:
\[ -\frac{1}{4} = -\frac{V}{U} \]
This simplifies to:
\[ \frac{V}{U} = \frac{1}{4} \]
From this, we can express \( V \) in terms of \( U \):
\[ V = \frac{U}{4} \]

### Step 3: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{V} + \frac{1}{U} \]
Substituting \( f = -12 \, \text{cm} \) and \( V = \frac{U}{4} \):
\[ \frac{1}{-12} = \frac{1}{\frac{U}{4}} + \frac{1}{U} \]
This simplifies to:
\[ \frac{1}{-12} = \frac{4}{U} + \frac{1}{U} \]
Combining the terms on the right side:
\[ \frac{1}{-12} = \frac{5}{U} \]

### Step 4: Solve for \( U \)
Cross-multiplying gives:
\[ 5 = -12U \]
Thus,
\[ U = -\frac{5}{12} \, \text{cm} \]
However, we need to correct the signs and ensure we solve correctly:
\[ U = -60 \, \text{cm} \]

### Conclusion
The object should be placed at a distance of \( 60 \, \text{cm} \) in front of the concave mirror.

### Final Answer
The object should be placed at \( 60 \, \text{cm} \) from the mirror.

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