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Calculate the power and focal length of the given lens. |
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Answer» \(\frac{1}{f}=\frac{1}{-20}-\frac{1}{-40}\) \(=+\frac{1}{20}+\frac{1}{40}\) \(=\frac{+2+1}{40}\) \(=\frac{+3}{40}\) \(f=\frac{+40}{3}\) \(P=\cfrac{100}{\frac{40}{3}}=\frac{300}{40}=\frac{30}{4}=7.5\) |
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