Two lenses are kept in contact one having focal length +40 cm and the other of focal length-20 cm.

(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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