Question

An object is placed at a distance of 18 cm in front of a mirror. If the image is formed at a distance of 4 cm on the other side, then the focal length, nature of the mirror and nature of the image are respectively:

• A. \(3.14 \, \text{cm}, \text{ concave mirror, and real image}\)
• B. \(3.14 \, \text{cm}, \text{ convex mirror, and real image}\)
• C. \(5.14 \, \text{cm}, \text{ convex mirror, and virtual image}\)
• D. \(5.14 \, \text{cm}, \text{ concave mirror, and virtual image}\)

Hint:

Remember the sign conventions in mirror equations: distances are negative if on the same side as the incoming light (real objects) and positive if on the opposite side (virtual images).

Answer 1

The Correct Option is C

Using the mirror equation: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] where \( u = -18 \, {cm} \) (object distance, negative for real object), and \( v = 4 \, {cm} \) (image distance, positive indicating virtual image). Solving for the focal length \( f \): \[ \frac{1}{f} = \frac{1}{-18} + \frac{1}{4} = -\frac{1}{18} + \frac{1}{4} = \frac{-1+4.5}{18} = \frac{3.5}{18} = \frac{7}{36} \] \[ f = \frac{36}{7} \approx 5.14 \, {cm} \] Since the image distance is positive and less than the object distance, the mirror is convex, and the image formed is virtual.