A candle is kept at a distance equal to double the focal length from the pole of a convex mirror. Its magnification will be :

Options

A. `-1//3`
B. `1//3`
C. `2//3`
D. `-2//3`

Answer

To solve the problem of finding the magnification of a candle placed at a distance equal to double the focal length from the pole of a convex mirror, we can follow these steps:

### Step 1: Understand the given information
- A candle (object) is placed at a distance of \(2F\) from the convex mirror.
- We need to find the magnification of the image formed by the mirror.

### Step 2: Set up the mirror formula
The mirror formula is given by:
\[ \frac{1}{V} + \frac{1}{U} = \frac{1}{F} \]
Where:
- \(V\) = image distance
- \(U\) = object distance
- \(F\) = focal length of the mirror

### Step 3: Assign the values according to the sign convention
- For a convex mirror, the focal length \(F\) is positive.
- The object distance \(U\) is negative (since it is measured against the direction of incident light). Given that the object is at a distance of \(2F\):
\[ U = -2F \]
- The focal length \(F\) is:
\[ F = +F \]

### Step 4: Substitute the values into the mirror formula
Substituting the values of \(U\) and \(F\) into the mirror formula:
\[ \frac{1}{V} + \frac{1}{-2F} = \frac{1}{F} \]

### Step 5: Solve for \(V\)
Rearranging the equation:
\[ \frac{1}{V} = \frac{1}{F} + \frac{1}{2F} \]
Finding a common denominator:
\[ \frac{1}{V} = \frac{2}{2F} + \frac{1}{2F} = \frac{3}{2F} \]
Now, taking the reciprocal to find \(V\):
\[ V = \frac{2F}{3} \]

### Step 6: Calculate the magnification
The magnification \(m\) is given by the formula:
\[ m = -\frac{V}{U} \]
Substituting the values of \(V\) and \(U\):
\[ m = -\frac{\frac{2F}{3}}{-2F} \]
This simplifies to:
\[ m = \frac{2F}{3} \cdot \frac{1}{2F} = \frac{1}{3} \]

### Step 7: Conclusion
The magnification of the image is:
\[ m = +\frac{1}{3} \]
This indicates that the image is virtual, erect, and diminished.