A convex lens of focal length 0.5 m and a co | vedclass

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(A) The power $P$ of a lens is given by $P = \frac{1}{f}$,where $f$ is the focal length in meters.For a convex lens,the focal length is positive,so $f

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(A) The power $P$ of a lens is given by $P = \frac{1}{f}$,where $f$ is the focal length in meters.For a convex lens,the focal length is positive,so $f_1 = 0.5 \ m$. Its power is $P_1 = \frac{1}{0.5} = 2 \ D$.For a concave lens,the focal length is negative,so $f_2 = -1 \ m$. Its power is $P_2 = \frac{1}{-1} = -1 \ D$.When lenses are combined,the total power is the sum of individual powers: $P = P_1 + P_2$.$P = 2 \ D + (-1 \ D) = 1 \ D$.

$A$ convex lens of focal length $0.5 \ m$ and a concave lens of focal length $1 \ m$ are combined. The power of the resulting lens will be: (in $D$)

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