A convex lens of focal length 25cm and a concave lens of focal length 10cm are placed in close contact with each other. Calculate the lens power of this combination.
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Focal length of convex lens $f_1=+25 cm=+0.25 m$ Power of convex lens $p_1$ $=\frac{1}{\text{f}_1}=\frac{1}{0.25}=4\text{D}$Focal length of concave lens $f_2$ $=-10\text{cm}=-0.10\text{m}$
Power of concave lens $p_2$ $=\frac{1}{\text{f}_2}=\frac{1}{-0.10}=-10\text{D}$
The power of the combination of lenses is the algebraic sum of the powers of the individual lenses.$\therefore$ Power of combination $P = P _1+ P _2$
⇒ P = 4 - 10 = - 6 D. The power of a lens and the focal length are related as$\text{p}=\frac{1}{\text{f}}$
$-6=\frac{1}{\text{f}}$
$\text{f}=\frac{1}{-6}=-0.167\text{m}-16.7\text{m}$
Therefore, the focal length of the combination of the lenses is -16.7cm.
Power of concave lens $p_2$ $=\frac{1}{\text{f}_2}=\frac{1}{-0.10}=-10\text{D}$
The power of the combination of lenses is the algebraic sum of the powers of the individual lenses.$\therefore$ Power of combination $P = P _1+ P _2$
⇒ P = 4 - 10 = - 6 D. The power of a lens and the focal length are related as$\text{p}=\frac{1}{\text{f}}$
$-6=\frac{1}{\text{f}}$
$\text{f}=\frac{1}{-6}=-0.167\text{m}-16.7\text{m}$
Therefore, the focal length of the combination of the lenses is -16.7cm.
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