Question

Using the mirror equation and the formula of magnification, deduce that “the virtual image produced by a convex mirror is always diminished in size and is located between the pole and the focus.”

Hint:

Convex mirrors always produce virtual, erect, and diminished images because the reflected rays diverge and appear to come from a point behind the mirror.

Answer 1

Mirror Equation: \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \] Sign Conventions for Convex Mirror: - Object is always placed in front of the mirror, so \( u<0 \) - Focal length \( f>0 \) (convex mirror) - Image formed is always virtual, upright, and on the same side as the object, so \( v>0 \) Step 1: Analyze using the Mirror Equation
Let’s assume \( f = +f \), \( u = -u \) \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{|u|} \Rightarrow v = \left( \frac{1}{f} + \frac{1}{|u|} \right)^{-1} \] Since both terms on the right are positive, \( \frac{1}{v}>\frac{1}{f} \Rightarrow v<f \) Hence, the image is located between the pole and the focus. Step 2: Magnification Formula
\[ m = \frac{h'}{h} = \frac{-v}{u} \] Since \( v>0 \), \( u<0 \Rightarrow m = \frac{-v}{- |u|} = \frac{v}{|u|} \) Since \( v<u \), this implies \( m<1 \Rightarrow \) image is diminished in size. Conclusion:
- The image is virtual (positive \( v \)), - Erect (positive magnification), - Diminished (magnification \(<1 \)), - Located between pole and focus (\( v<f \)). % Final Answer Statement Answer: Using the mirror equation and magnification formula, we conclude that a convex mirror always forms a virtual, erect, and diminished image located between the pole and the focus.