Question

Two thin lenses, when in contact, produce a combination of power $+10 \,D$. When they are $0.25\, m$ apart, the power is reduced to $+6 \,D$. The power of the lenses in dioptres, are

• A. 1 and 9
• B. 2 and 8
• C. 4 and 6
• D. 5 each

Answer 1

Answer: (B)

Case I When lenses are in contact, $P=\frac{1}{F_{1}}=+10 D$
$\frac{1}{F_{1}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$ \Rightarrow 10=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
Case II When lenses are $d=0.25 m$ apart
$P=\frac{1}{F_{2}}=+6 D$
$ \Rightarrow \frac{1}{F_{2}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}} \ldots (ii) $
From Eq. (i),
$ 6=10-\frac{0.25}{f_{1} f_{2}} $
or $ f_{1} f_{2}=\frac{0.25}{4}=\frac{1}{16}$
$\Rightarrow \frac{1}{f_{1}} \times \frac{1}{f_{2}} =16 $
$\therefore P_{1} P_{2} =16\,\,\,...(iii)$
Hence, from Eq. (iii), only option (b) satisfies this relation.
Hence $P_{1}$ and $P_{0}$ is $2$ and $8$ .