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Difficulty: MediumReflection of Light at Plane and Curved Mirrors

A convex spherical mirror always forms a virtual, erect, and diminished image of a real object, regardless of the object's distance from the mirror.

Answer: Answer

Answer

The statement is True. A convex mirror consistently produces virtual, erect, and diminished images for all real object positions.
The statement is correct because the outward curvature of a convex mirror causes all incident parallel or diverging rays from a real object to diverge upon reflection. The virtual extensions of these rays converge behind the mirror between the pole and the focus, ensuring the image is always virtual, upright, and smaller than the object.

Step-by-Step Solution

1
Identify the sign conventions for a convex mirror and a real object.
The focal length ff is negative (f<0f < 0) because the focus is behind the mirror, and the object distance uu is positive (u>0u > 0) for a real object.
Establishing proper sign convention is essential for analyzing image formation in curved mirrors.
2
Analyze the mirror equation to determine the sign of the image distance vv.
From 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}, re-arranging gives 1v=1f−1u=−(1∣f∣+1u)\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\left(\frac{1}{|f|} + \frac{1}{u}\right). Thus, vv is always negative.
A negative image distance (v<0v < 0) mathematically proves that the image is virtual and located behind the mirror.
3
Evaluate linear magnification mm to determine image orientation and size.
Using m=−vum = -\frac{v}{u}, since v<0v < 0 and u>0u > 0, m>0m > 0 (erect image). Additionally, ∣v∣=∣f∣uu+∣f∣<u|v| = \frac{|f|u}{u + |f|} < u, so ∣m∣=∣v∣u<1|m| = \frac{|v|}{u} < 1 (diminished image).
Magnification sign indicates orientation (positive is erect) and magnitude indicates size relative to the object.

Key Concept

Image characteristics in convex mirrors
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Topics

Reflection of Light at Plane and Curved Mirrors

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