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