Question

The magnification of the virtual image of an object placed at a distance of 0.2 m from the lens is 0.5. The lens will be:

• A. A concave lens of focal length 1 m.
• B. A concave lens of focal length 0.2 m.
• C. A convex lens of focal length 0.1 m.
• D. A convex lens of focal length 0.2 m.

Hint:

Concave lenses always form virtual, upright, and diminished images, whereas convex lenses can form both real and virtual images.

Answer 1

The Correct Option is B

Given: - The object distance \( u = -0.2 \, \text{m} \) (negative because the object is placed on the same side as the incident light).
- The magnification \( M = 0.5 \).
- The image is virtual, which implies a positive magnification for a concave lens.
Step 1: Use the magnification formula
The magnification \( M \) for a lens is given by the formula: \[ M = \frac{v}{u} \] Where:
- \( v \) is the image distance.
- \( u \) is the object distance.
Substitute the known values: \[ 0.5 = \frac{v}{-0.2} \] Solving for \( v \): \[ v = 0.5 \times (-0.2) = -0.1 \, \text{m} \] Thus, the image distance \( v = -0.1 \, \text{m} \). The negative sign indicates that the image is virtual and formed on the same side as the object (characteristic of concave lenses).
Step 2: Use the lens formula to find the focal length The lens formula is: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Where:
- \( f \) is the focal length of the lens.
- \( v \) is the image distance.
- \( u \) is the object distance.
Substitute the known values \( v = -0.1 \, \text{m} \) and \( u = -0.2 \, \text{m} \): \[ \frac{1}{f} = \frac{1}{-0.1} - \frac{1}{-0.2} \] \[ \frac{1}{f} = -10 + 5 = -5 \] Therefore: \[ f = \frac{1}{-5} = -0.2 \, \text{m} \] The negative focal length indicates that the lens is a concave lens.
Step 3: Conclusion
Thus, the lens is a concave lens with a focal length of \( 0.2 \, \text{m} \).
The correct answer is: \[ \boxed{(B) A concave lens of focal length 0.2 m.} \]