Question

The distance (v) of a virtual image formed by a lens of focal length 15 cm can never exceed a certain finite value, then this value will be:

• A. less than 15 cm
• B. between 15 cm to 30 cm
• C. less than or equal to 30 cm
• D. less than or equal to 15 cm

Hint:

A key property to remember: A concave lens always forms a virtual, erect, and diminished image, located between the focus and the optical center.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the limit on the distance of a virtual image formed by a lens. A convex lens can form a virtual image whose distance can be very large (approaching infinity). However, a concave lens always forms a virtual image within a specific range. Given the options, the question is likely about a concave lens.
Step 2: Key Formula or Approach:
We will analyze the image formation by a concave lens using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] For a concave lens, the focal length (\(f\)) is negative. By convention, the object distance (\(u\)) is also negative. The image formed is always virtual, so the image distance (\(v\)) will also be negative.
Step 3: Detailed Explanation:
Let the focal length be \(f = -15\) cm.
The lens formula is \(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\).
Substituting \(f = -15\):
\[ \frac{1}{v} = \frac{1}{-15} + \frac{1}{u} \] Since the object is real, \(u\) can range from 0 to \(-\infty\).
Case 1: Object is at infinity (\(u \rightarrow -\infty\)).
\[ \frac{1}{v} = \frac{1}{-15} + \frac{1}{-\infty} = -\frac{1}{15} + 0 \] \[ v = -15 \text{ cm} \] This means the image is formed at the focus. The image distance is 15 cm.
Case 2: Object is at the optical center (\(u \rightarrow 0\)).
\[ \frac{1}{v} = \frac{1}{-15} + \frac{1}{0} \rightarrow \infty \] This limit is not practical, but as the object moves from infinity towards the lens, the image moves from the focus towards the optical center.
For any real object placed in front of a concave lens, the virtual image is always formed between the optical center (O) and the principal focus (F).
Therefore, the image distance |v| will always be less than the focal length |f|.
So, \(0<|v| \leq |f|\).
Given \(f=15\) cm, the image distance \(v\) will be less than or equal to 15 cm.
Step 4: Final Answer:
For a concave lens of focal length 15 cm, the virtual image is always formed between the optical center and the focus, so its distance from the lens can never exceed 15 cm.