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Interference after reflection from thin plate

1 vote

I don't understand the solution given. It describes this situation as identical to a YDSE setup where distance between the slits =2d and the nearest slit is at a distance of 2l from the screen.(also attached figure).Could someone please explain how they are equivalent?

I thought the path difference(of lamda/2) due to reflection of the two days from the two different sides of the plate P gives rise to interference.

asked Dec 3, 2021 in Physics Problems by thewitness (110 points)
Please can you post the text which accompanies the 2nd diagram (ie the solution)?

The 2nd diagram is confusing. The question says "fringes of equal inclination." This means that the rays which interfere should be travelling parallel towards the screen at the same angle $\theta$. This is not the case in the diagram.

1 Answer

0 votes

The first diagram shows all rays emerging from a single source S and being reflected from the plate back to the screen. The second diagram shows the same rays "unfolded" as though they are coming from 2 "virtual sources" S1 and S2 behind the plate similar to "virtual images" behind a mirror.

Why are there 2 virtual sources? This is because rays travel a different optical path length OPL from the pinhole source S back to screen, depending on whether they are reflected from the front or rear face of the plate. The extra OPL travelled by rays reflected close to the normal from the back face is approx. $2nd$ where $n$ is the refractive index of the plate. So the distances of virtual sources S1 and S2 from the screen are $2\ell$ and $2\ell+2nd$. Not quite as given in the diagram.

Waves from sources S1 and S2 are emitted out of phase by $\pi$ radians because there is a change of phase on reflection from the front face of the plate but not from the rear face. So rays emitted from S1 and S2 at angle $\theta$ to the axis will be in phase if the difference in their OPL (which is S2N = S1S2$\cos\theta$ in the 2nd diagram) is equal to an odd integer $2k-1$ of half-wavelengths, :
$$2nd\cos\theta=\frac12\lambda (2k-1)$$

For small angles $\theta$ we can use the approximations $\cos\theta \approx 1-\frac12 \theta^2$ and $\theta \approx \frac{r}{2\ell}=\frac{D}{4\ell}$ where $D$ is the diameter of a fringe, which is a ring. Setting $\frac{\lambda}{2nd}=x$ we can write
$$2-\theta^2 \approx (2k-1)x$$

We can find $\lambda$ by plotting a graph. Measure the diameters $D$ of fringes of order $k=1, 2, 3$ etc. Calculate corresponding values of $\theta$. Plot values of $\theta^2$ against corresponding values $2k-1$. This should give a straight line of slope $-x$ from which $\lambda$ can be calculated - assuming that we know the values of $n, d$.

answered Feb 19 by sammy gerbil (28,876 points)
edited Feb 20 by sammy gerbil