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How will the light pass through the combination of convex and concave lens of power equal in magnitude? |
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Answer» Power of convex lens, P1 = \(\frac{1}{f_1}\) Power of concave lens, P2 = \(\frac{1}{-f_2}\) Combination will have power, P = P1 + P2 = \(\frac{1}{f_1}+\frac{1}{-f_2}\) Given that = \(\frac{1}{f_1}=\frac{1}{f_2}\) or P1 = P2 ∴ P = \(\frac{1}{f_1}+\frac{1}{-f_2}\) = 0 light will pass undeviated |
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