Question

An object is placed at the focus of concave lens having focal length f. What is the magnification and distance of the image from the optical centre of the lens ?

• A. Very high, \( \infty \)
• B. 1, \( \infty \)
• C. \( \frac{1}{2}, \frac{f}{2} \)
• D. \( \frac{1}{4}, \frac{f}{4} \)

Hint:

Remember that a concave lens always forms a virtual, erect, and diminished image. Unlike a convex lens, placing an object at the focus of a concave lens does not create an image at infinity.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a concave lens, the focal length \( f \) is always negative according to the sign convention.
The object is placed at the focus, so \( u = -f \) (where \( f \) is the magnitude of focal length).
Step 2: Key Formula or Approach:
1. Lens Formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \).
2. Magnification: \( m = \frac{v}{u} \).
Step 3: Detailed Explanation:
Given \( u = -f \) and \( F = -f \):
Substitute into the lens formula:
\[ \frac{1}{v} - \frac{1}{-f} = \frac{1}{-f} \]
\[ \frac{1}{v} + \frac{1}{f} = -\frac{1}{f} \]
\[ \frac{1}{v} = -\frac{1}{f} - \frac{1}{f} = -\frac{2}{f} \]
\[ v = -\frac{f}{2} \]
The image distance from the optical center is \( \frac{f}{2} \).
Now, calculate magnification:
\[ m = \frac{v}{u} = \frac{-f/2}{-f} = \frac{1}{2} \]
Step 4: Final Answer:
The magnification is \( 1/2 \) and the image distance is \( f/2 \).