An object is placed at 5 cm distance in front of concave mirror of radius of curvature 15 cm.
What is the nature and magnification of the image?
Answer & explanation
Correct answer: option 4
u = -5cm , f = -7.5 cm
$ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{-1}{7.5} + \frac{1}{5} = \frac{1}{15}$
$ v = 15cm$
Since v is positive , so image is virtual.
$m = -\frac{v}{u} = +3$
SO image is erect.