The dispersive powers of glasses of lenses used in an achromatic pair are in the ratio 5 : 3. If the focal length of the concave lens is 15 cm , then the nature and focal length of the other lens would be （ ）
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The dispersive powers of glasses of lenses used in an achromatic pair are in the ratio 5 : 3. If the focal length of the concave lens is 15 cm , then the nature and focal length of the other lens would be （ ）
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Solution
Given: The ratio of dispersive powers, \(\dfrac{{\mathit{\omega }}_{1}}{{\mathit{\omega }}_{2}}=\dfrac{5}{3}\)
Focal length of the concave lens, \({f}_{1}=15cm\)
Using the condition, \(\dfrac{{\omega }_{1}}{{f}_{1}}+\dfrac{{\omega }_{2}}{{f}_{2}}=0\)
\(\dfrac{{\omega }_{1}}{{f}_{1}}=−\dfrac{{\omega }_{2}}{{f}_{2}}\)
\(\dfrac{{\omega }_{1}}{{\omega }_{2}}=−\dfrac{{f}_{1}}{{f}_{2}}\)
\(\dfrac{5}{3}=−\dfrac{-15}{{f}_{2}}\)
\({f}_{2}=9\;cm\)
Positive sign indicates the lens is convex in nature.
Focal length of the concave lens, \({f}_{1}=15cm\)
Using the condition, \(\dfrac{{\omega }_{1}}{{f}_{1}}+\dfrac{{\omega }_{2}}{{f}_{2}}=0\)
\(\dfrac{{\omega }_{1}}{{f}_{1}}=−\dfrac{{\omega }_{2}}{{f}_{2}}\)
\(\dfrac{{\omega }_{1}}{{\omega }_{2}}=−\dfrac{{f}_{1}}{{f}_{2}}\)
\(\dfrac{5}{3}=−\dfrac{-15}{{f}_{2}}\)
\({f}_{2}=9\;cm\)
Positive sign indicates the lens is convex in nature.
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