Two thin lenses have a combined power of +9 D. When they are separated by a distance of 20 cm , their…
Two thin lenses have a combined power of +9 D. When they are separated by a distance of 20 cm , their equivalent power becomes $+\frac{27}{5} \mathrm{D}$. The power of both the lenses in dioptre are respectively
4,5
3,6
2,7
1,8
Solution
$\begin{aligned}
& P_1+P_2=9 D \ldots . \text { (i) } \\
& P_1+P_2-d_1 P_2=\frac{27}{5} \ldots . \text { (ii) }
\end{aligned}$
where $\mathrm{d}=20 \mathrm{~cm}=0.2 \mathrm{~m}$
Substituting value of $\left(\mathrm{P}_1+\mathrm{P}_2\right)$ in equation (ii) we get
$\begin{array}{ll}
& 9-0.2 P_1 P_2=\frac{27}{5} \\
\therefore & 0.2 P_1 P_2=9-\frac{27}{5}=\frac{18}{5} \\
\therefore & P_1 P_2=\frac{18}{0.2 \times 5}=18
\end{array}$
Since $P_1+P_2=9$ and $P_1 P_2=18$
$\therefore \quad \mathrm{P}_1=3$ and $\mathrm{P}_2=6$
.