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

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,
\(
\begin{aligned}
\mathrm{P}_1 & =8 \mathrm{D} \\
\therefore \quad \mathrm{P}_2 & =10-8=2 \mathrm{D}
\end{aligned}
\)