Question

An object of height 3.6 cm is placed normally on the principal axis of a concave mirror of radius of curvature 30 cm. If the object is at a distance of 10 cm from the principal focus of the mirror, then the height of the real image formed due to the mirror is

• A. 5.4 cm
• B. 3.6 cm
• C. 1.8 cm
• D. 2.7 cm

Hint:

Always draw a rough ray diagram for mirror/lens problems. For a concave mirror, an object placed between C (-30 cm) and F (-15 cm), like at u=-25 cm, will form a real, inverted, and magnified image beyond C. This helps in verifying the signs and relative magnitude of your answer.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires the use of the mirror formula to find the image position and the magnification formula to find the image height. It's crucial to use the correct sign convention for all quantities. For a concave mirror, the focal length and radius of curvature are negative. Distances measured in the direction of incident light are positive, and those opposite are negative.
Step 2: Key Formula or Approach:
- Mirror Formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
- Magnification: \(m = \frac{h_i}{h_o} = -\frac{v}{u}\)
where \(f\) is focal length, \(v\) is image distance, \(u\) is object distance, \(h_i\) is image height, and \(h_o\) is object height.
Step 3: Detailed Explanation:
Determine the parameters from the question:
- Object height, \(h_o = +3.6\) cm (placed normally, so considered upright).
- Radius of curvature, \(R = -30\) cm (concave mirror).
- Focal length, \(f = R/2 = -30/2 = -15\) cm.
- The object is "10 cm from the principal focus". The focus is at x = -15 cm. For a real image to be formed by a concave mirror, the object must be placed beyond the focus. Therefore, the object position is farther from the mirror than the focus.
- Object distance, \(u = -15 \text{ cm} - 10 \text{ cm} = -25\) cm.
Calculate the image distance (v):
Using the mirror formula:
\[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-25} \] \[ \frac{1}{v} = \frac{1}{25} - \frac{1}{15} = \frac{3 - 5}{75} = \frac{-2}{75} \] \[ v = -\frac{75}{2} = -37.5 \text{ cm} \] The negative sign indicates that the image is formed on the same side as the object and is therefore a real image, which is consistent with the question.
Calculate the image height (\(h_i\)):
Using the magnification formula:
\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \] \[ \frac{h_i}{3.6} = -\frac{-37.5}{-25} = -1.5 \] \[ h_i = -1.5 \times 3.6 = -5.4 \text{ cm} \] The negative sign indicates the image is inverted. The height (magnitude) of the image is 5.4 cm.
Step 4: Final Answer:
The height of the real image is 5.4 cm. Therefore, option (A) is correct.