Lateral shift of a glass slab nd refraction by a prism
Answers
In terms of the incidence angle θ1, the slab thickness d and the refraction index n, I have found that the lateral shift x is given by
x=dn2sin(θ1)(n2+sin2(θ1)−−−−−−−−−−√)(n2−sin2(θ1)−−−−−−−−−−√−cos(θ1)).
In order to test this formula I have considered two extreme cases: when the light is perpendicular to the slab (θ1=0) and when it is almost parallel (θ1≈π/2). The first case works OK, since if θ1=0 then sin(θ1)=0 and x=0. This is physically plausible, for if the light ray is perpendicular then it does not change direction.
I'm having trouble assessing the second case. If θ1≈π/2 then the formula can be written as
x=dn2n4−1−−−−−√.
Is this physically plausible? My guess has been that if I can say that n4≫1 I obtain x≈d. This sounds correct to me since if the light ray approaches the slab almost parallel then after refracting back to air it should come out almost parallel, causing a lateral shift of exactly d. I am not sure, however, in saying n4≫1. I have checked glass's refraction index and it lies in the range 1.5−1.9. Am I correct?