The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X be 15 cm :(a) calculate the focal length of lens Y.(b) state the nature of lens Y.

(a) Given:

Focal length of lens X, $f_X$ = $+$15 cm = $+$0.15 m

Power of the combination of lenses X and Y, $P_X+P_Y$ = $+$5 D

To find: Focal length of lens Y, $f_Y$

Solution:

Power of the lens is given by-

$P=\frac {1}{f}$

Substituting the given value we get-

$P_X=\frac {1}{0.15}$

$P_X=\frac {100}{15}$

$P_X=+6.67D$

Now, putting the value of $P_X$ in the expression of combination of lenses.

$P_X+P_Y=+5D$

$6.67+P_Y=5$

$P_Y=5-6.67$

$P_Y=-1.67D$

The focal length of the lens is related to the power of the lens as:

$f=\frac {1}{P}$

therefore, the focal length of the lens Y $(f_Y)$ is given as-

$f_Y=\frac {1}{P_Y}$

$f_Y=\frac {1}{-1.67}$

$f_Y=-\frac {100}{167}$ 

$f_Y=-0.60m=-60cm$ 

Thus, the focal length of the lens Y $(f_Y)$ is -60cm.

(b) The nature of the lens Y is diverging, i.e, concave since it has a negative focal length.

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Updated on: 10-Oct-2022

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