1.

How will the light pass through the combination of convex and concave lens of power equal in magnitude?

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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