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: Note the object sits between $f$ and $2f$.
With $f=10$ cm and object at $15$ cm, the object lies between the focus and twice the focus, so we expect a real, inverted, magnified image.
Step 2: Use the magnification-from-distances shortcut.
For a thin lens the linear magnification can be written $m = \dfrac{f}{f+u}$ using the real-is-positive convention with object distance magnitude.
Step 3: Plug values in.
Taking $u=-15$ cm, $f=+10$ cm: \[ m = \frac{f}{f+u} = \frac{10}{10+(-15)} = \frac{10}{-5} = -2 \]
Step 4: Read off the sign.
The negative sign tells us the image is inverted, and the magnitude $2$ tells us it is twice as tall.
Step 5: Scale the object height.
\[ |h_i| = |m|\,h_o = 2 \times 4 = 8\ \text{cm} \]
Step 6: State the physical image.
The image is real, inverted, and $8$ cm tall, matching the key without ever solving for $v$ separately. \[ \boxed{h_i = 8\ \text{cm (inverted)}} \]