A dentist concave mirror has a radius of curvature of `30 cm`. How far must it be placed from a small cavity in order to get a virtual image magnified `5` times ?

Explanation

To solve the problem, we need to find out how far the dentist should place the concave mirror from the small cavity so that a virtual image is formed that is magnified 5 times. Here are the steps to arrive at the solution:

### Step 1: Determine the Focal Length
The radius of curvature (R) of the concave mirror is given as 30 cm. The focal length (f) of a concave mirror is given by the formula:
\[ f = \frac{R}{2} \]
Substituting the value of R:
\[ f = \frac{30 \, \text{cm}}{2} = 15 \, \text{cm} \]

### Step 2: Use the Magnification Formula
The magnification (m) for mirrors is given by the formula:
\[ m = -\frac{v}{u} \]
Where:
- \( v \) = image distance
- \( u \) = object distance (which is negative for real objects)

Given that the magnification is 5 (and since it’s a virtual image, it's positive), we can write:
\[ 5 = -\frac{v}{u} \]
This implies:
\[ v = -5u \]

### Step 3: Apply the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting \( f = -15 \, \text{cm} \) (focal length is negative for concave mirrors):
\[ -\frac{1}{15} = \frac{1}{v} + \frac{1}{u} \]

### Step 4: Substitute for v
From the magnification equation, we have \( v = -5u \). Substituting this into the mirror formula:
\[ -\frac{1}{15} = \frac{1}{-5u} + \frac{1}{u} \]

### Step 5: Simplify the Equation
Finding a common denominator for the right side:
\[ -\frac{1}{15} = -\frac{1}{5u} + \frac{5}{5u} \]
This simplifies to:
\[ -\frac{1}{15} = \frac{4}{5u} \]

### Step 6: Cross-Multiply to Solve for u
Cross-multiplying gives:
\[ -1 \cdot 5u = 4 \cdot -15 \]
Thus:
\[ -5u = -60 \]
Now, dividing both sides by -5:
\[ u = 12 \, \text{cm} \]

### Conclusion
The object (small cavity) must be placed **12 cm** from the concave mirror.

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