An object is placed 10 cm in front of a convex lens of focal length 15 cm. The image produced will be
Real and magnified
Virtual and magnified
Virtual and reduced in size
Real and reduced in size
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-1 – 2015
$1/15 = 1/v – 1/(-10)$
$1/15 = 1/v + 1/10$
$1/v = 1/15 – 1/10$
$1/v = (2 – 3)/30$
$1/v = -1/30$
$v = -30$ cm
The image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image.
The magnification ($m$) is given by $m = v/u$:
$m = (-30) / (-10) = +3$
The positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object).
Therefore, the image is virtual and magnified.