- #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 √(n

^{2}sin

^{2}θ - 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 √(n

^{2}sin

^{2}θ - n'

^{2}) = 1

√(n

^{2}sin

^{2}θ - n'

^{2}) = 1 / 2*π

(n

^{2}sin

^{2}θ - n'

^{2}) = √ (1 / 2*π)

n

^{2}sin

^{2}θ = √ (1 / 2*π) + n'

^{2}

sin

^{2}θ = [ √ (1 / 2*π) + 12 ] / 1.332

sin

^{2}θ = 0.79085

√sin

^{2}θ = √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 sin

^{2}(62.78°) - 1

^{2}) ] = 1.386 * 10

^{-7}m

So the decay length of 1.386 * 10

^{-7}m is not equal to the length of the evanescent wave of 5.5*10

^{-7}m.

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