Phase shifted waves due to path difference

[kankerfist]

Homework Statement

A wave front moving horizontally encounters a block of glass that has an index of refraction of 1.50 in its upper half and an index of refraction of 1.25 in its lower half. The wavelength of the wave front in air is 700 nm. At what lengths would the glass block cause constructive interference (bright spots) and destructive interference (dark spots)?

Homework Equations

n1=1.50
n2=1.25
λair=700E-9 m

The Attempt at a Solution

I came up with the following when I encountered this study problem and I was hoping someone could double check to see if I did this correctly:

Constructive interference should occur when the glass block length causes a phase shift of (integer)(2pi). When 2 rays leave the block they will be phase shifted by (2pi)(length of block)/(λ in n1 and n2). So the total phase shift for 2 rays going through the block:

(2pi)(L)/(λ in n1) - (2pi)(L)/(λ in n2)

and:
λ in n1= λ in air / n1 = (700E-9)/(1.5)
λ in n2 = (700E-9)/(1.25)

so constructive interference should occur when:

(2pi)(L) / [(700E-9)/(1.5)] -
(2pi)(L) / [(700E-9)/(1.25)] =
(integer)(2pi)

this reduces to:

Length of glass = 7(integer)/2.5E+6

So constructive interference should occur when the length of the glass block is any positive integer times (7/2.5*10^6) in meters. If anybody could take a glance and see if I messed up somewhere i'd appreciate it!

 

[AvroArrow]

For destructive interference, I assumed it should occur when the total phase shift is an odd multiple of pi. The same equations would apply except now: Length of glass = 7(odd integer)/2.5E+6 Thank you!

 

[m2003]

Your approach and calculations seem correct to me. When a wave passes through a medium with a different refractive index, there will be a phase shift due to the path difference between the two rays. As you correctly stated, constructive interference occurs when the phase shift is a multiple of 2pi, and this will happen at lengths of the glass block that are integer multiples of (7/2.5*10^6) meters. Similarly, destructive interference will occur when the phase shift is an odd multiple of pi. Good job on your solution!

 

[kyle8921]

Your approach to solving this problem is correct. The key concept to understand is that the path difference between the two rays is responsible for the phase shift and hence the interference pattern. Constructive interference occurs when the path difference is an integer multiple of the wavelength, while destructive interference occurs when the path difference is an odd multiple of half the wavelength. Your calculations show that the length of the glass block should be an integer multiple of (7/2.5*10^6) meters for constructive interference to occur. This is consistent with the general formula for path difference, which is given by (n2-n1)L/λ, where n1 and n2 are the indices of refraction and L is the length of the medium. Keep up the good work!