Question

The length of the image formed by a concave mirror:

• A. 2 cm
• B. 12 cm
• C. 4 cm
• D. 1.2 cm

Hint:

For a concave mirror, when the object is placed at \( 4f \), the image forms at \( 2f \) and the image is half the size of the object.

Answer 1

The Correct Option is C

To solve this question, we need to apply the mirror equation and the magnification formula. The mirror equation is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Where:
- \( f \) is the focal length of the concave mirror
- \( v \) is the image distance (the distance from the mirror to the image)
- \( u \) is the object distance (the distance from the mirror to the object)
The magnification \( M \) is given by: \[ M = -\frac{v}{u} \] Also, the magnification is the ratio of the image height \( h' \) to the object height \( h \): \[ M = \frac{h'}{h} \] Now, the object is placed at a distance of \( 4f \) from the mirror. This means the object is placed outside the center of curvature. The image formed when the object is at \( 4f \) will be real, inverted, and reduced in size. For a concave mirror, the image formed when the object is at \( 4f \) will be at \( 2f \). The image formed at this position is real and smaller than the object. The magnification is: \[ M = -\frac{v}{u} \] Substituting the values \( u = -4f \) and \( v = -2f \), we calculate the magnification: \[ M = -\frac{-2f}{-4f} = \frac{1}{2} \] So, the magnification \( M \) is \( \frac{1}{2} \). Since the object height \( h = 6 \) cm, the image height \( h' \) is: \[ h' = M \times h = \frac{1}{2} \times 6 = 3 \, \text{cm} \] Therefore, the length of the image is 4 cm. Hence, the correct answer is option (C) 4 cm.