First slide

Ray optics

Question

Match the corresponding entries of column 1 with column 2.[Where $m$ is the magnification produced by the mirror]

Moderate

A

$A \to pand s; B \to q and r; C \to q and s; D \to q and r$
B

$A \to r and s; B \to q and s; C \to q and r; D \to p and s$
C

$A \to q and r; B \to q and r; C \to q and s; D \to p and s$
D

$A \to pand r; B \to pand s; C \to pand q; D \to r and s$

Solution

$Magnification in the mirror, m = - \frac{v}{u}$
A) $m = - 2 \Rightarrow v = 2 u$
$A s v a n d u h a v e s a m e s i g n s s o t h e m i r r o r i s c o n c a v e a n d i m a g e f o r m e d i s r e a l .$
B) $m = - \frac{1}{2} \Rightarrow v = \frac{u}{2} \Rightarrow Concave mirror and real image.$
C) $m = + 2 \Rightarrow v = - 2 u$
$A s v a n d u h a v e d i f f e r e n t s i g n s b u t m a g n i f i c a t i o n i s 2 s o t h e m i r r o r i s c o n c a v e a n d i m a g e f o r m e d i s v i r t u a l .$
D) $m = + \frac{1}{2} \Rightarrow v = - \frac{u}{2}$
As v and u have different signs with magnification $(\frac{1}{2})$ so the mirror is convex and image formed is virtual.

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