(i) \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) ( is negative)

\(u=-f⇒\frac{1}{v}=0⇒v=∞\)

\(u=-2f⇒\frac{1}{v}=\frac{-1}{2f}⇒v=-2f\)

Hence if -2f < u < -f

⇒ -2f < u < ∞

(ii) \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

Using sign convention, for convex mirror, we have

f > 0, u < 0

From the formula

\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

∵ f is positive and u is negative,

⇒ v is always positive, hence image is always virtual.

(iii) Using mirror formula

\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

For concave mirror, f < 0

\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

For an object placed between focal and pole of mirror; f < u < 0

Hence \(\frac{1}{u}<\frac{1}{f}\) . It means \(\frac{1}{v}\) > 0

Hence image is always virtual.

Also, as f < 0

\(\frac{1}{v}-\frac{1}{u}<0\)

\(\frac{1}{v}-\frac{1}{u}<0\) (v is always positive)

∵ u < v

So, \(m=|\frac{v}{u}|>1,\)

Hence, image is always enlarged.