A pencil is placed upright at a distance of 10 cm from a convex lens of focal length 15 cm. The nature of the image of the pencil will be
[amp_mcq option1=”real, inverted and magnified” option2=”real, erect and magnified” option3=”virtual, erect and reduced” option4=”virtual, erect and magnified” correct=”option4″]
This question was previously asked in
UPSC NDA-1 – 2016
For a convex lens, when the object is placed between the optical centre and the principal focus (i.e., 0 < |u| < f, or 0 < 10 cm < 15 cm), the image formed is virtual, erect, and magnified.
– Object between optical centre and F: Image is virtual, erect, magnified, and on the same side as the object.
1/15 = 1/v – 1/(-10)
1/15 = 1/v + 1/10
1/v = 1/15 – 1/10 = (2 – 3)/30 = -1/30
v = -30 cm.
Since v is negative, the image is virtual and formed on the same side as the object.
Magnification m = v/u = -30 / -10 = +3.
Since m is positive, the image is erect. Since |m| > 1, the image is magnified.
Thus, the image is virtual, erect, and magnified.