An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre of the lens is 20 cm. Position, nature and size of the image formed.

Options

A. 6.67 cm, -1.67 cm, real, inverted
B. -6.67 cm, 1.67 cm, virtual, erect
C. -6.67 cm, -1.67 cm, real, erect
D. -6.67 cm, -1.67 cm, real, inverted

Answer

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

### Step 1: Identify the given values
- Height of the object (h_o) = 5 cm
- Focal length of the concave lens (f) = -10 cm (negative because it is a concave lens)
- Object distance (u) = -20 cm (negative according to the sign convention, as the object is on the same side as the incoming light)

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) = focal length of the lens
- \( v \) = image distance from the lens
- \( u \) = object distance from the lens

Substituting the known values into the lens formula:
\[ \frac{1}{-10} = \frac{1}{v} - \frac{1}{-20} \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{-10} + \frac{1}{20} \]

### Step 4: Finding a common denominator
To add the fractions, we find a common denominator (which is 20):
\[ \frac{1}{-10} = \frac{-2}{20} \]
Thus,
\[ \frac{1}{v} = \frac{-2}{20} + \frac{1}{20} = \frac{-1}{20} \]

### Step 5: Solving for \( v \)
Now, taking the reciprocal to find \( v \):
\[ v = -20 \, \text{cm} \]

### Step 6: Determine the nature of the image
Since \( v \) is negative, the image is formed on the same side as the object, indicating that the image is virtual.

### Step 7: Calculate the magnification
The magnification (m) is given by the formula:
\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \]
Substituting the known values:
\[ m = -\frac{-20}{-20} = -1 \]

### Step 8: Finding the height of the image
Using the magnification to find the height of the image \( h_i \):
\[ h_i = m \cdot h_o = -1 \cdot 5 = -5 \, \text{cm} \]

### Step 9: Conclusion
- **Position of the image**: 20 cm from the lens on the same side as the object.
- **Nature of the image**: Virtual and erect (since the height is negative, it indicates that the image is erect).
- **Size of the image**: The height of the image is 5 cm.

### Summary of Results
- Position: 20 cm from the lens (same side as the object)
- Nature: Virtual and erect
- Size: 5 cm