Explanation

(d) For ${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V}_{i} < C$
the diode is reverse biased and hence offer infinite resistance, so circuit would be like as shown in Fig. (2) and $V_{o} = 0.$
[image:14_image205.emf]
For $V_{i} > 0,$the diode is forward biased and circuit would be as shown in Fig. (3) and $V_{o} = V_{i}$.
Hence, the optical (d) is correct.