Decay length of the evanescent wave

  • #1
594
12

Homework Statement


A beam of light of vacuum wavelength λ = 550nm passes from water (refractive index 1.33) into air (refractive index 1.00).

(a) What is the critical angle?

(b) Suppose the beam is totally internally reflected. At what angle of incidence would the decay length of the evanescent wave be equal to λ?

Homework Equations


δ = λ / [ 2*π √(n2 sin2θ - n'2) ]

where

λ = 550 nm

δ: decay length

n = 1.33

n'= 1.00

The Attempt at a Solution



a) = 48.75° - That's pretty simple to calculate the critical angle

I'm having a few issues with b) though.

δ = λ, therefore:

2*π √(n2 sin2θ - n'2) = 1

√(n2 sin2θ - n'2) = 1 / 2*π

(n2 sin2θ - n'2) = √ (1 / 2*π)

n2 sin2θ = √ (1 / 2*π) + n'2

sin2θ = [ √ (1 / 2*π) + 12 ] / 1.332

sin2θ = 0.79085

√sin2θ = √0.79085

sinθ = 0.88929747553

sin-1(0.88929747553) = 62.78°

So the angle of incidence I get equals 62.78°. When I plug this back into the original equation though I get:

(5.5*10-7) / [ 2*π √(1.332 sin2(62.78°) - 12) ] = 1.386 * 10-7m

So the decay length of 1.386 * 10-7m is not equal to the length of the evanescent wave of 5.5*10-7m.

I'm not sure where I've gone wrong here. Any help would be much appreciated.
 

Answers and Replies

  • #2
35,919
12,746
√(n2 sin2θ - n'2) = 1 / 2*π

(n2 sin2θ - n'2) = √ (1 / 2*π)

n2 sin2θ = √ (1 / 2*π) + n'2

sin2θ = [ √ (1 / 2*π) + 12 ] / 1.332
You might want to check these steps.
 
  • #3
594
12
√(n2 sin2θ - n'2) = 1 / 2*π

(n2 sin2θ - n'2) = 1 / 4π2

(n2 sin2θ) = (1 / 4π2) + n'2

sin2θ = ( (1 / 4π2) + n'2 ) / n2

√ sin2θ = √ [ ( (1 / 4π2) + n'2 ) / n2 ]

sinθ = √ [ ( (1 / 4π2) + 1.002 ) / 1.332 ]

sinθ = 0.76134

sin-1θ = 49.58°

so...

(5.5*10-7) / [ 2*π √(1.332 sin2(49.58°) - 12) ] = 5.5*10-7m

Thanks @mfb - can't believe I didn't see that initial mistake! Must have been looking at the problem for too long.
 

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