Finding the ratio of thicknesses of films (destructive)

[nso09]

Homework Statement

I need help on figuring out part C which is determining the ratio between $t_1$ and $t_2$ in order to get constructive interference. I'm given the index of refractions $n_1=1.5$ and $n_2=2$ and that I know that wave 1 and 2 have half cycle shift and wave 3 does not.
I only really need help with part c since I figured out the rest.

Homework Equations

1. $2t=(m+.5)λ$
2. $2t=(m)λ$

The Attempt at a Solution

I basically did the same thing for how I did part b. Since both ray 1 and 2 have shifts, I used the 2nd equation. And for ray 2 and 3, I used the 1st equation since there is a pi difference.
$$2t_1=(m+.5)(\lambda/n_1)$$ and set m =0 since we want minimum t thickness.
$$2t_2=m(\lambda/n_2)$$ and set m =1 since we want minimum thickness also.
so
$$2t_1=\lambda/(2(n_1))$$
$$2t_2=\lambda/n_2$$

then I divide both $2t_1$ and $2t_2$ so I have this:
$t_1/t_2=n_2/2n_1$ since the 2s and $\lambda$ cancel out.
Since $n_1=1.5$ and $n_2=2$,
$$t_1/t_2=2/2(1.5)$$ therefore $t_1/t_2=2/3$
However, the solutions say it is $t_1/t_2=2n_2/5n_1=2*2/5*1.5=0.5333$
I don't get how I did the same method for constructive yet can't use it for destructive.

 

[TSny]

The equation that you set up for rays 1 and 2 would make these two rays destroy each other. If that happens, then what would be left to give destructive interference for ray 3?

You will need to consider the condition for 3 waves to cancel, rather than the condition for two waves to cancel. If you have studied phasors then they can be helpful.
See figure 37.12 here http://www.kshitij-iitjee.com/phasor-diagrams-of-waves

[The question apparently wants you to neglect the change in amplitude of the reflected and the transmitted wave at an interface between two different materials. Taking this into account would probably change the answer significantly.]

 

[nso09]

The equation that you set up for rays 1 and 2 would make these two rays destroy each other. If that happens, then what would be left to give destructive interference for ray 3?

You will need to consider the condition for 3 waves to cancel, rather than the condition for two waves to cancel. If you have studied phasors then they can be helpful.
See figure 37.12 here http://www.kshitij-iitjee.com/phasor-diagrams-of-waves

[The question apparently wants you to neglect the change in amplitude of the reflected and the transmitted wave at an interface between two different materials. Taking this into account would probably change the answer significantly.]

I know three waves/sources would have $\phi=2\pi/3$ so I basically have a triangle that cancels itself out.
What I don't understand is when I looked at the solutions, it said that $$ 4\pi(t_1n_1)/\lambda=2\pi(m)+2\pi/3 $$
and $$ 4\pi(t_2n_2)/\lambda -\pi=2\pi(m)+2\pi/3 $$
Why is there a pi in the second equation? Is it because on the paths of wave 1 and wave 2, there is no relative phase shift but there is one between wave 2 and 3?
I think I get how they got the two equations because it's basically the wave number times the path difference equaling a full wavelength $\2pi$ times an integer.

 

[TSny]

I know three waves/sources would have $\phi=2\pi/3$ so I basically have a triangle that cancels itself out.

Yes.

Why is there a pi in the second equation? Is it because on the paths of wave 1 and wave 2, there is no relative phase shift but there is one between wave 2 and 3?

Yes.

 

[nso09]

Yes.
Yes.

Okay thank you. You mentioned earlier that my first attempt at a solution can't work because it made the first two waves cancel therefore there is no other wave to work with wave 3. But this time, the thin film equation is taking into account the phase difference $2\pi/3$ to make ALL THREE WAVES cancel which is why this works.