Two lenses A and B have focal lengths of +20 cm and, −10 cm, respectively.
(a) What is the nature of lens A and lens B?
(b) What is the power of lens A and lens B?
(c) What is the power of combination if lenses A and B are held close together?
(a) The nature of lens A is convex because it has a positive focal length, i.e $+$20 cm.
The nature of lens B is concave because it has a negative focal length, i.e $-$10cm.
(b) Given:
Focal length of lens A, $f_A$ = $+$20 cm = $+$0.2 m
Focal length of lens A, $f_B$ = $-$10 cm = $-$0.1 m
To find: Power of lens A, $(P_A)$ and lens B, $(P_B)$.
Solution:
Power of the lens is given by-
$P=\frac {1}{f}$
Substituting the given values we get-
$P_A=\frac {1}{0.2}=\frac {10}{2}=+5D$
$P_B=\frac {1}{-0.1}=-\frac {10}{1}=-10D$
Thus, the power of lens A, $(P_A)$ and lens B, $(P_B)$ is +5D and -10D respectively.
(c) The power of combination, if lenses A and B are held close together, is given as-
$P=P_A+P_B=5D+(-10D)=5D-10D=-5D$
Thus, the power of the combination of lenses A and B is -5D.
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