Interference on a mirror -- waves confusion

[nso09]

Homework Statement

$d=0.52 cm,$ $\lambda=431nm$, $L=119 m$

Homework Equations

$dsin\theta=(m+1/2)\lambda$ for constructive/bright fringes since there is a pi shift already

The Attempt at a Solution

a) $$sin\theta\leq1$$
$$(.0052/431e-9)-0.5=m=12064$$ 2 \e{3}
$$ total bright fringes = 2m = 24129$$

However, I am extremely confused on why my solutions sets $d=.0052(2)$ which gives $m=24129$ from the get go.

Isn't there a mirror underneath? I don't know where this solution is coming from which is messing up my numbers for parts b and c. Is there something I'm missing? The main thing that I don't get is why $d=.0104m$

 

[Simon Bridge]

The point source S is distance L from the screen and distance d above the mirror.
The interference is between this source and it's image S' in the mirror ... what is the separation between the source and it's image?

 

[nso09]

The point source S is distance L from the screen and distance d above the mirror.
The interference is between this source and it's image S' in the mirror ... what is the separation between the source and it's image?

Oh ok. separation between source and image is .0104m. But why do we have to take into account the image in the mirror? I'm not sure how that's relevant in the problem

 

[Simon Bridge]

Oh ok. separation between source and image is .0104m. But why do we have to take into account the image in the mirror? I'm not sure how that's relevant in the problem

The image in a mirror is the place where light rays appear to radiate from ... this means the image acts like another source.
So we can replace the physical setup with one that does not have the mirror, but there is a second source identical with the first but with opposite phase... this new setup will produce the exact same interference pattern as the one with the mirror.
Then do the maths for this setup - remember that there are no fringes in the lower half though.

You asked why d=0.0104m i the solution - and that is because the d in $d\sin\theta =(m+\frac{1}{2})\lambda$ is not the same as the d in the diagram.
0.0104m is the separation between S and it's image.

You don't have to do it that way though - you can just work out the equation for the path difference and derive the result.

 

[nso09]

The image in a mirror is the place where light rays appear to radiate from ... this means the image acts like another source.
So we can replace the physical setup with one that does not have the mirror, but there is a second source identical with the first but with opposite phase... this new setup will produce the exact same interference pattern as the one with the mirror.
Then do the maths for this setup - remember that there are no fringes in the lower half though.

You asked why d=0.0104m i the solution - and that is because the d in $d\sin\theta =(m+\frac{1}{2})\lambda$ is not the same as the d in the diagram.
0.0104m is the separation between S and it's image.

You don't have to do it that way though - you can just work out the equation for the path difference and derive the result.

I see. So this is just a slightly different type of double slit problem in which the equations are switched between destructive and constructive because there is a pi shift in one of the sources, providing a relative pi phase shift.

 

[Simon Bridge]

I see. So this is just a slightly different type of double slit problem in which the equations are switched between destructive and constructive because there is a pi shift in one of the sources, providing a relative pi phase shift.

Well done ... it's the equivalent to "method of images" in electrostatics.

The trick to these sorts of things is to be careful about deriving the equations you use.
If you can replace the setup you have with an equivalent that you already have the answers to, then so much the better.
Why do more maths than you have to?

 

[nso09]

Well done ... it's the equivalent to "method of images" in electrostatics.

The trick to these sorts of things is to be careful about deriving the equations you use.
If you can replace the setup you have with an equivalent that you already have the answers to, then so much the better.
Why do more maths than you have to?

That's true. Thank you so much for helping me out.