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How does the lateral dispacment depend upon angle of incidence |
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Answer» Yes it does By using elementary geomety, it can be proved that d1d2=cosicosr where d1 and d2 are the width of the light beam in the incident and refracted medium respectively, and i and r are the incident and refracted angle respectively Lateral Shift would be ∣d1−d2∣ or Suppose your slab has a certain thickness, say d. From Snell's LAW about REFRACTION, calling θi the angle between the incoming ray and the normal to the surface, ni the refractive index of the medium outside the slab, NM the refractive index INSIDE the slab and θm the angle between the ray in the slab and the normal to the surface: nmsinθm=nisinθi you can derive the angle θm: sinθm=ninmsinθi Then, from basic trigonometry, you know that: sin2θm+cos2θm=1 ⇒cosθm=1−sin2θθm−−−−−−−−−−√ and, with dd as defined and l length of the PATH of the ray inside the slab, you can write: d=l⋅cosθm ⇒l=dcosθm Calling x the lateral displacement you are asking for: x=l⋅sinθm By substituting in this equation, you can get your result as an expression of the angle of incidence, the refractive indexes and the thickness of the slab. |
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