The correct option is (a) 2
For explanation: Given: f = +10 cm; v = -30 cm
Required equations ➔
\(\frac {1}{f}=\frac {1}{v}-\frac {1}{u}\)
m=\(\frac {size \, of \, the \, image}{size \, of \, the \, object}=\frac {v}{u}\)
\(\frac {1}{u}=\frac {1}{v}-\frac {1}{f}=\frac {1}{-30}-\frac {1}{10} = \frac {-(10+30)}{300} = \frac {-40}{300}= \frac {-2}{15}\)
u=\(\frac {-15}{2}\) = -7.5 cm
m=\(\frac {v}{u}= – \frac {30}{-7.5}=\frac {-300}{-75}\)= +4
Therefore, the object should be placed at a distance of 7.5 cm from the lens to get the image at a distance of 30 cm from the lens. It is four times enlarged and is erect.