how do you find the lateral shift of a glass slab???????????
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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)=sinθ1cosθ2−cosθ1sinθ2dL=cosθ2
Assuming that the air has a refractive index of 1, we can further write
sinθ1sinθ2=n
sinθ1sinθ2=n
From basic geometry we know that for angles in the first quadrant,
cosθ=1−sin2θ−−−−−−−−√
cosθ=1−sin2θ
Combining these gives
x=dcosθ2(sinθ1cosθ2−cosθ1sinθ2)=d(sinθ1−sinθ1cosθ1ncosθ2)=dsinθ1⎛⎝⎜1−1−sin2θ1−−−−−−−−√n1−sin2θ1n2−−−−−−−−√⎞⎠⎟=dsinθ1(1−1−sin2θ1−−−−−−−−√n2−sin2θ1−−−−−−−−−√)
x=dcosθ2(sinθ1cosθ2−cosθ1sinθ2)=d(sinθ1−sinθ1cosθ1ncosθ2)=dsinθ1(1−1−sin2θ1n1−sin2θ1n2)=dsinθ1(1−1−sin2θ1n2−sin2θ1)
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...
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)=sinθ1cosθ2−cosθ1sinθ2dL=cosθ2
Assuming that the air has a refractive index of 1, we can further write
sinθ1sinθ2=n
sinθ1sinθ2=n
From basic geometry we know that for angles in the first quadrant,
cosθ=1−sin2θ−−−−−−−−√
cosθ=1−sin2θ
Combining these gives
x=dcosθ2(sinθ1cosθ2−cosθ1sinθ2)=d(sinθ1−sinθ1cosθ1ncosθ2)=dsinθ1⎛⎝⎜1−1−sin2θ1−−−−−−−−√n1−sin2θ1n2−−−−−−−−√⎞⎠⎟=dsinθ1(1−1−sin2θ1−−−−−−−−√n2−sin2θ1−−−−−−−−−√)
x=dcosθ2(sinθ1cosθ2−cosθ1sinθ2)=d(sinθ1−sinθ1cosθ1ncosθ2)=dsinθ1(1−1−sin2θ1n1−sin2θ1n2)=dsinθ1(1−1−sin2θ1n2−sin2θ1)
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...
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