Question

When an object is kept at distance 8 cm and 24 cm from a convex lens, magnitude of magnification is same in both cases. Find focal length (in cm) of the lens :

• A. 18 cm
• B. 16 cm
• C. 20 cm
• D. 8 cm

Hint:

This is a standard problem type. The condition of equal magnification magnitude for two different object positions with a convex lens usually implies one real and one virtual image, leading to the relation \(m_1 = -m_2\).
Remembering the magnification formula \(m = f/(f+u)\) can be faster than using the lens formula and \(m=v/u\) separately.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a convex lens. When an object is placed at two different positions, \(u_1 = 8\) cm and \(u_2 = 24\) cm, the magnitude of the magnification is the same. We need to find the focal length (\(f\)) of the lens.
Step 2: Key Formula or Approach:
The magnification (\(m\)) produced by a lens is given by the formula:
\[ m = \frac{f}{f+u} \] where \(f\) is the focal length and \(u\) is the object distance. We must use the sign convention, so object distances will be negative.
Step 3: Detailed Explanation:
According to the sign convention, the object is placed to the left of the lens, so the object distances are negative.
Case 1: \(u_1 = -8\) cm.
Case 2: \(u_2 = -24\) cm.
We are given that the magnitudes of the magnifications are equal: \(|m_1| = |m_2|\).
For a convex lens, if the object is placed between the optical center and the focus (\(|u|<f\)), the image is virtual and erect (\(m>0\)). If the object is placed beyond the focus (\(|u|>f\)), the image is real and inverted (\(m<0\)).
Since we have two different positions giving the same magnification magnitude, it's highly probable that one position is within the focal length and the other is outside, meaning one image is virtual and the other is real. Thus, we have \(m_1 = -m_2\).
Using the magnification formula for both cases:
\[ m_1 = \frac{f}{f + u_1} = \frac{f}{f - 8} \] \[ m_2 = \frac{f}{f + u_2} = \frac{f}{f - 24} \] Now, applying the condition \(m_1 = -m_2\):
\[ \frac{f}{f - 8} = - \left( \frac{f}{f - 24} \right) \] Since \(f \neq 0\), we can cancel \(f\) from both sides:
\[ \frac{1}{f - 8} = \frac{-1}{f - 24} \] Cross-multiplying gives:
\[ f - 24 = -(f - 8) \] \[ f - 24 = -f + 8 \] \[ 2f = 32 \] \[ f = 16 \text{ cm} \] Step 4: Final Answer:
The focal length of the convex lens is 16 cm. This confirms our assumption: \(u_1 = 8\) cm is within the focus, giving a virtual image, and \(u_2 = 24\) cm is beyond the focus, giving a real image.