Question

how do you find the lateral shift of a glass slab???????????

Answer 1

I get a different formula. Let me show you how I derived it.

Using the following diagram:

enter image description here

We can write the following equations by looking at triangles:

xLdL=sin(θ1−θ2)=sinθ1cosθ2−cosθ1sinθ2=cosθ2
xL=sin⁡(θ1−θ2)=sin⁡θ1cos⁡θ2−cos⁡θ1sin⁡θ2dL=cos⁡θ2
Assuming that the air has a refractive index of 1, we can further write

sinθ1sinθ2=n
sin⁡θ1sin⁡θ2=n
From basic geometry we know that for angles in the first quadrant,

cosθ=1−sin2θ−−−−−−−−√
cos⁡θ=1−sin2⁡θ
Combining these gives

x=dcosθ2(sinθ1cosθ2−cosθ1sinθ2)=d(sinθ1−sinθ1cosθ1ncosθ2)=dsinθ1⎛⎝⎜1−1−sin2θ1−−−−−−−−√n1−sin2θ1n2−−−−−−−−√⎞⎠⎟=dsinθ1(1−1−sin2θ1−−−−−−−−√n2−sin2θ1−−−−−−−−−√)
x=dcos⁡θ2(sin⁡θ1cos⁡θ2−cos⁡θ1sin⁡θ2)=d(sin⁡θ1−sin⁡θ1cos⁡θ1ncos⁡θ2)=dsin⁡θ1(1−1−sin2⁡θ1n1−sin2⁡θ1n2)=dsin⁡θ1(1−1−sin2⁡θ1n2−sin2⁡θ1)
Note that with this expression, the distance xx will approach dd when θ1θ1 approaches π/2π/2 since the second term will vanish.

You might want to compare my approach with yours. I'm not claiming mine is right...