This type of question is based on the concept of the diode. As we know that the in forward bias, a p-type semiconductor is at a higher potential as compared to an n-type semiconductor. So by using this concept we will be able to answer this question.Complete Step By Step Solution:
In forward bias, the $p$ side of $p - n$ the junction is connected to higher potential, and $n$ the side of $p - n$ the junction is connected to lower potential. So we can say that the potential will be higher than another side.
Now we will look into the figure,
As we can see in the option $\left( b \right)$ the potential on the left side is much higher than the potential on the right side. So we can say that this diode represents the forward bias diode.Therefore, the option $\left( b \right)$ is correct.Additional information:
Forward bias in the electronics world is when, in a SEMICONDUCTOR device, when we connect its $P - type$ , positive doped region, i.e. with B, Al, etc. to the positive terminal of the battery and $N - type$ , Negative doped I.e. with N, P, etc. to the negative terminal of the battery.
Forward bias, the immediate current-voltage is needed to keep up the current stream in a bipolar semiconductor or diode or to improve the current stream in a field-impact semiconductor. A silicon diode will lead current just if its anode is at a positive voltage contrasted with its cathode; it is then supposed to be forward one-sided.Note:
As we can see that this question becomes very easy when we know about the properties or we can say the definition of it. So we should have to understand the concept and memorize the properties of bias to answer the question.