A beam of light converges towards a point O, behind a convex mirror of focal length 20cm.
Q. Find the magnification and nature of the image when point O is 10cm behind the mirror.

Options

A. `2`(virtual,inverted)
B. `3`(real, inverted)
C. `5`(virtual, enlarged)
D. `2` (real, erect)

Explanation

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

### Step 1: Identify the given values
- Focal length of the convex mirror (F) = +20 cm (positive for convex mirrors)
- Distance of the point O behind the mirror (u) = -10 cm (negative because the object is virtual and located behind the mirror)

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values into the formula:
\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-10} \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{20} + \frac{1}{10} \]
To add these fractions, we need a common denominator. The common denominator for 20 and 10 is 20:
\[ \frac{1}{v} = \frac{1}{20} + \frac{2}{20} = \frac{3}{20} \]

### Step 4: Solve for v
Now, taking the reciprocal to find v:
\[ v = \frac{20}{3} \approx 6.67 \text{ cm} \]
Since v is positive, this indicates that the image is formed on the same side as the object (in front of the mirror).

### Step 5: Find the magnification (m)
The magnification (m) is given by the formula:
\[ m = -\frac{v}{u} \]
Substituting the values we found:
\[ m = -\frac{6.67}{-10} = \frac{6.67}{10} = 0.667 \]

### Step 6: Determine the nature of the image
Since the magnification is positive, the image is erect. The value of magnification being less than 1 indicates that the image is smaller than the object.

### Final Results
- The image distance (v) is approximately 6.67 cm in front of the mirror.
- The magnification (m) is approximately 0.667, indicating that the image is erect and smaller than the object.