# What is the minimum value of L for which the waves will undergo full destructive interference?

Two light waves travel in air with wavelength 520 nm, and they are initially in phase. Each passes through a block of

length L before passing back into the air. One travels through glass with index of refraction n1 = 1.65, and the other

passes through ice with index of refraction n2 = 1.31. What is the minimum value of L for which the waves will undergo

full destructive interference?

My Analysis :

For destructive interference there should be a phase shift. But in this case since air has n=1 and it is going to another medium, there is already a phase shift.

So we can use the equation ; dsin(t) = m*lambda/n

n1 = 1.65

using m=1

dsin(t) = 520nm/1.65

dsin(t) = 315.15nm

This value is not accepted as the right answer, also since theta is not known, it is hard to find the distance or length d. For total internal reflection we can use sin(90) , so I am not clear about the approach to solve this problem. Please help.

### 1 Answer

- NCSLv 78 months agoFavorite Answer
I would assume that the angle of incidence for both beams is 0º.

The wavelength in the first medium is

520nm / 1.65 = 315.15 nm

and in the second medium is

520nm / 1.31 = 396.95 nm

For a block of length L, we'll get a number of wavelengths

n = L / λ

and so

L = nλ

We need n+½ wavelengths of the slower wave and n wavelengths of the faster one in order to get destructive interference, so

L = (n+½)*315.15 = n*396.95

solves to

n = 1.93

L = 1.93*396.95nm = 766 nm

Yes?