Question

An object of height \( 4 \, \text{cm} \) is placed \( 15 \, \text{cm} \) in front of a convex lens of focal length \( 10 \, \text{cm} \). The height of the image formed is:

• A. \( 4 \, \text{cm} \)
• B. \( 6 \, \text{cm} \)
• C. \( 8 \, \text{cm} \)
• D. \( 12 \, \text{cm} \)

Hint:

For a convex lens, if the object is placed between \( f \) and \( 2f \) (here, \( 10 \, \text{cm} < 15 \, \text{cm} < 20 \, \text{cm} \)), the image formed is real, inverted, magnified, and formed beyond \( 2f \). Since it is magnified, the height must be greater than \( 4 \, \text{cm} \), ruling out option (A) immediately.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
An object of height \( 4 \, \text{cm} \) is kept in front of a convex lens. We need to find the height of the formed image.

Step 2: Key Formula or Approach:
Use the lens formula to find the image distance \( v \):
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Then, use the magnification formula to find the height of the image \( h_i \):
\[ m = \frac{v}{u} = \frac{h_i}{h_o} \]

Step 3: Detailed Explanation:
Given:
- Object height, \( h_o = 4 \, \text{cm} \)
- Object distance, \( u = -15 \, \text{cm} \) (using coordinate sign convention)
- Focal length of convex lens, \( f = +10 \, \text{cm} \)
Substitute \( f \) and \( u \) into the lens formula:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-15} \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \] \[ \frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30} \implies v = 30 \, \text{cm} \] Now, determine the magnification \( m \):
\[ m = \frac{v}{u} = \frac{30}{-15} = -2 \] Since \( m = \frac{h_i}{h_o} \):
\[ -2 = \frac{h_i}{4} \implies h_i = -8 \, \text{cm} \] The negative sign indicates that the image is inverted. The magnitude of the height of the image is \( 8 \, \text{cm} \).

Step 4: Final Answer:
(C) \( 8 \, \text{cm} \)