A combination of two thin lenses in contact have power $+10 \mathrm{D}$. The power reduces to $+6…
A combination of two thin lenses in contact have power $+10 \mathrm{D}$. The power reduces to $+6 \mathrm{D}$ when the lenses are $0.25 \mathrm{~m}$ apart. The power of individual lens is
$5 \mathrm{D}, 5 \mathrm{D}$
$6 \mathrm{D}, 4 \mathrm{D}$
$7 \mathrm{D}, 3 \mathrm{D}$
$8 \mathrm{D}, 2 \mathrm{D}$
Solution
When in contact, the equivalent power of two thin lenses is,
$\mathrm{P}_1+\mathrm{P}_2=10$
When the lenses are separated by $0.25 \mathrm{~m}$, $\mathrm{P}_1+\mathrm{P}_2-0.25 \mathrm{P}_1 \mathrm{P}_2=6$
$\begin{array}{ll}
\therefore & 0.25 \mathrm{P}_1 \mathrm{P}_2=4 \\
\therefore & \mathrm{P}_1 \mathrm{P}_2=16 \\
& \mathrm{P}_1-\mathrm{P}_2=\sqrt{\left(\mathrm{P}_1+\mathrm{P}_2\right)^2+4 \mathrm{P}_1 \mathrm{P}_2} \\
\therefore & \mathrm{P}_1-\mathrm{P}_2=\sqrt{36}=6...(i)
\end{array}$
Adding equations (i) and (ii), we get,
$\begin{aligned}
\mathrm{P}_1 & =8 \mathrm{D} \\
\therefore \quad \mathrm{P}_2 & =10-8=2 \mathrm{D}
\end{aligned}$