Two lenses of power -15 D and +5 D are in c | vedclass

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(C) The power of the combination of lenses in contact is given by the algebraic sum of their individual powers:$P = P_{1} + P_{2}$Given $P_{1} = -15 \

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$-10$

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(C) The power of the combination of lenses in contact is given by the algebraic sum of their individual powers:$P = P_{1} + P_{2}$Given $P_{1} = -15 \ D$ and $P_{2} = +5 \ D$,we have:$P = (-15 + 5) \ D = -10 \ D$Since the power $P$ is related to the focal length $f$ (in meters) by the formula $P = \frac{1}{f}$,we can find the focal length as:$f = \frac{1}{P} = \frac{1}{-10} \ m$To convert the focal length into centimeters,we multiply by $100$:$f = \left( \frac{1}{-10} 	imes 100 ight) \ cm = -10 \ cm$Thus,the focal length of the combination is $-10 \ cm$.

Two lenses of power $-15 \ D$ and $+5 \ D$ are in contact with each other. The focal length of the combination is ....... $cm$.

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