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Obtian the equation for lateral displacement of light passing through a glass slab. |
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Answer» Solution :Consider a glass slab of thickness t and refractive index n is KEPT in AIR medium. The path of the light is ABCD and the refractions occur at two points B and C in the galss slab. The angles of incidence I and refraction r are measured with respect to the normal `N_(1) and N_(2)` at the twopoints B and C respectively. The lateral displacement L is the perpendicular distance CE drawnbetween the path of light and the undeviated path of light at point C. In the right agle triangle `Delta BCE`, `sin(i-r)=(L)/(BC),BC=(L)/(sin(i-r))` In the right angle triangle `Delta BCF`, `cos(r)=(t)/(BC),BC=(t)/(cos(r))` Equating equations (1) and (2) `(L)/(sin(i-r))=(t)/(cos(r))` After rearranging, `L=t(sin(i-rr)/(cos(r)))` Lateral displacement depends upon the thickness of the slab. Thickher the slab, greater will be the lateral displacement. Greater the angle of incident, larger will be the lateral displacement. 
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