What is the relative decrease in focal length | vedclass

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(A) The initial optical power is $P = 2.5 \ D$. The initial focal length is $F = \frac{1}{P} = \frac{1}{2.5} = 0.4 \ m$.When the power increases by $0

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$0.04$

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(A) The initial optical power is $P = 2.5 \ D$. The initial focal length is $F = \frac{1}{P} = \frac{1}{2.5} = 0.4 \ m$.When the power increases by $0.1 \ D$,the new power is $P' = 2.5 + 0.1 = 2.6 \ D$.The new focal length is $F' = \frac{1}{P'} = \frac{1}{2.6} \ m$.The relative decrease in focal length is given by $\frac{F - F'}{F} = 1 - \frac{F'}{F}$.Substituting the values,we get $1 - \frac{1/2.6}{1/2.5} = 1 - \frac{2.5}{2.6} = 1 - \frac{25}{26} = \frac{1}{26}$.Calculating the value,$\frac{1}{26} \approx 0.03846$,which rounds to $0.04$.

What is the relative decrease in focal length of a lens for an increase in optical power by $0.1 \ D$ from $2.5 \ D$? [$D$ stands for dioptre]

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