Question

Two lenses of power $+10 \, D$ and $-5 \, D$ are placed in contact. Where should an object be held from the lens, so as to obtain a virtual image of magnification $2$ ?

• A. $5 \, cm$
• B. $-5$
• C. $10 \, cm$
• D. $-10 \, cm$

Answer 1

Answer: (D)

Here, $P_{1}=+10 \, D, \, P_{2}=-5 \, D$
Combined power,
$P=P_{1}+P_{2}=10 \, D-5 \, D$
$P=5 \, D$
$\therefore $ Combined focal length
$=\frac{1}{P} \, m=\frac{100}{P} \, cm=\frac{100}{5}=20 \, cm$
Given magnification,
$m=2$
$\frac{v}{u}=2$
$v=2u$
By lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{2 u}-\frac{1}{u}=\frac{1}{20}$
$\frac{1 - 2}{2 u}=\frac{1}{20}$
$2u=20$
$u=-10 \, cm$
Hence, to obtain virtual image, object should be place at a distance of 10 cm