# Why in this problem the phase shift due to reflection from wall is not considered? [closed]

I know in reflection from rigid surface the path difference occurring is $$\lambda/2$$; where $$\lambda$$ is wavelength. But in this problem they did not consider that. Is the solution to this problem correct?

• It is not possible to understand from the graphic you show just what you are asking. However, in general phase shift of a reflected wave depends upon the boundary conditions at the interface. Commented Jun 8, 2019 at 14:54
• Hi Anik and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. Commented Jun 8, 2019 at 14:58
• @John The OP is not asking for an answer to a homework question. In fact they have followed the instructons in the link you provided quite well as the question has been tagged "homework" and the OP has narrowed their question to one particular issue that is adjacent to the homework question. Could you please remove your comment as I do not believe it appropriate and could be taken as discouraging further posting by the OP. Commented Jun 8, 2019 at 15:18
• I have not asked any solution to this problem .I just want to know the solution is correct or not ?if it had been a homework question I could have it clarified from my tuition teacher . John Rennie have you gone through the problem .? Did it seem to you a homework question ? Well it's an illustrative example . I would request you to remove the homework exercise tag . And mind to answer to the point what I asked yet. Commented Jun 8, 2019 at 16:33
• Note that's questions of the form "Is this right" tend to be poor fits for this site because the answer is too short to be a valid answer Commented Jun 9, 2019 at 10:51

The solution is flawed, by more than just this. "The minimum distance x for which a maximum occurs at detector, the path length will be $$\Delta = \lambda$$" as a statement has not been justified and is completely untrue. A minimum for x requires a maximum for \Delta which you can see if you plot the two or differentiate. If we were to ignore the phase shift like is done in the answer this would give x=0 the minimal distance.