Calculate the distance of an object of height `h` from a concave mirror of focal length `10 cm` so as to obtain a real image of magnification `2`.

Explanation

To solve the problem, we need to find the distance of an object from a concave mirror given the focal length and the magnification. Let's break it down step by step.

### Step 1: Understand the given information
- Focal length of the concave mirror, \( f = -10 \, \text{cm} \) (negative because it is a concave mirror).
- Magnification \( m = -2 \) (since it is a real image, magnification is negative).

### Step 2: Relate magnification to object and image distances
The magnification \( m \) is given by the formula:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
Where:
- \( h' \) is the height of the image,
- \( h \) is the height of the object,
- \( v \) is the image distance,
- \( u \) is the object distance.

Since the magnification is \( -2 \):
\[ -2 = -\frac{v}{u} \implies v = 2u \]

### Step 3: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values we have:
\[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{u} \]

### Step 4: Substitute \( v \) in terms of \( u \)
From the magnification relation, we have \( v = 2u \). Now substitute this into the mirror formula:
\[ \frac{1}{-10} = \frac{1}{2u} + \frac{1}{u} \]

### Step 5: Simplify the equation
To combine the fractions on the right side, find a common denominator:
\[ \frac{1}{2u} + \frac{1}{u} = \frac{1 + 2}{2u} = \frac{3}{2u} \]
Now the equation becomes:
\[ \frac{1}{-10} = \frac{3}{2u} \]

### Step 6: Cross-multiply to solve for \( u \)
Cross-multiplying gives:
\[ -10 \cdot 3 = 2u \implies -30 = 2u \implies u = -15 \, \text{cm} \]
The negative sign indicates that the object is on the same side as the incoming light, which is consistent with the convention for concave mirrors.

### Final Answer
The distance of the object from the concave mirror is \( 15 \, \text{cm} \).

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