A 3.0 cm high object 5 cm from a convex mirror produces a virtual image. If the mirror has a focal length of 5 cm, what is height of the image?

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We’re given a 3.0 cm tall object placed 5 cm from a convex mirror with focal length 5 cm, and the image is virtual. We need to analyze each option and explain why the correct one fits the physics of convex mirrors, while the others do not.

Option a: 10 cm image height. If the image height were 10 cm, the magnification would be h_i/h_o = 10/3 ≈ 3.33. A magnification greater than 1 would imply the image is larger than the object. For a convex mirror, any real or virtual image that occurs is upright and smaller than the object (the image is always diminished, not magnified). Therefore, 10 cm is inconsistent with a convex mirror producing a virtual image. This option wrongly implies a magnification greater than 1 and a larger image, which is physically implausible for this setup.

Option b: 1 cm image height. This would give magnification m = h_i/h_o = 1/3 ≈ 0.333. While convex mirrors do produce upright, diminished images, the exact geometry (object distance 5 cm, f = 5 cm) leads to a smaller magnitude of image height that is proportional to the object, but the precise algebra for a convex mirror with these values yields a specific diminished height. A height of 1 cm would correspond to a magnification of about 0.333, which does not align with the standard result for these given optical parameters (the more typical diminished result here is 0.5 of the object height). In short, while being diminished is correct in spirit, the value 1 cm does not match the correct proportional reduction given the typical sign convention and distances for this problem.

Option c: 5 cm image height. This would give magnification m = h_i/h_o = 5/3 ≈ 1.667, meaning the image is larger than the object. As established for convex mirrors, images formed are smaller than the object (they are upright and diminished). This option contradicts the fundamental property of convex-mirror images and is thus incorrect.

Option d: 1.5 cm image height. This yields magnification m = h_i/h_o = 1.5/3 = 0.5, so the image is half the height of the object. For a convex mirror, the image is virtual, upright, and diminished. Using the sign convention consistent with a convex mirror and the given distances, the magnitude of the image distance v is 2.5 cm behind the mirror (virtual), which gives a magnification of |v|/u = 2.5/5 = 0.5. This produces the correct, smaller image height of 1.5 cm. Among the options, this one matches the expected qualitative behavior (virtual, upright, diminished) and the quantitative reduction by a factor of 2, making it consistent with the given data.

Summary of reasoning: A convex mirror with f = 5 cm produces a virtual, upright, diminished image for an object placed at 5 cm. The height reduction factor should be 0.5, giving h_i = 0.5 × h_o = 0.5 × 3.0 cm = 1.5 cm. The other options either imply a larger image, an incorrect amount of magnification, or an inverted/real-image scenario that contradicts the nature of a convex mirror in this setup.

A 3.0 cm high object 5 cm from a convex mirror produces a virtual image. If the mirror has a focal length of 5 cm, what is height of the image?

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