An object is placed 10 cm away from a convex lens of focal length 15 cm. The image produced is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → At 2F, virtual, 3 times magnified.
The object is between the optical centre and the focus of a convex lens, so the image is virtual, erect and magnified. The lens formula with the Cartesian sign convention (distances measured from the optical centre, positive in the direction of the incident light) confirms where it is.
\( u = -10\,\mathrm{cm} \)
\( f = +15\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \displaystyle \frac{1}{v} = \frac{1}{15} + \frac{1}{-10} \)
\( \displaystyle \frac{1}{v} = \frac{2 - 3}{30} = -\frac{1}{30} \)
\( v = -30\,\mathrm{cm} \)
The negative sign means the image is on the same side as the object, so it is virtual. Its distance, 30 cm, is twice the focal length, so it is at 2F.
\( \displaystyle m = \frac{v}{u} = \frac{-30}{-10} \)
\( m = +3 \)
The image is erect and 3 times magnified.