An object, 3.0 cm in height, is placed at a distance of 20.0 cm in front of a convex mirror of focal length 6.0 cm on its principal axis. Its image has a height of ________ and is ____________.

Correct option is D

​$$\text{Given Data:} \\h_o = 3.0 \, \text{cm}, \, u = -20.0 \, \text{cm}, \, f = 6.0 \, \text{cm} \\\text{Step 1: Use the Mirror Formula to Find Image Distance (\(v\))} \\\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\\text{Substitute the values:} \\\frac{1}{6} = \frac{1}{v} + \frac{1}{-20} \\\text{Simplify:} \\\frac{1}{v} = \frac{1}{6} + \frac{1}{20} \\\text{Find the common denominator:} \\\frac{1}{v} = \frac{10}{60} + \frac{3}{60} = \frac{13}{60} \\\text{Solve for \(v\):} \\v = \frac{60}{13} \approx 4.62 \, \text{cm} \\\text{Step 2: Use the Magnification Formula to Find Image Height (\(h_i\))} \\M = \frac{h_i}{h_o} = \frac{v}{u} \\\text{Substitute values:} \\M = \frac{4.62}{-20} \approx -0.231 \\\text{Solve for \(h_i\):} \\h_i = M \cdot h_o = (-0.231) \cdot 3.0 \approx -0.693 \, \text{cm} \\\text{Final Answer:} \\\text{The height of the image is:} \, h_i \approx 0.693 \, \text{cm} \\\text{The image is virtual, upright, and diminished.}$$​​