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A ray of light is incidentan angle of 50^(@) on one face of a cube of side 0.10 m. If the refractive index of the material of the glass cube is 1.55 then calculate the amount of lateral shift produced by it. |
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Answer» Solution :Given `N = 1.55 t = 0.10m i = 50^(@) L.S =` ? `LS=(t sin(i-r))/(COS r)""`.......(1) `n= (sin i)/(sin r)""`.......(2) `therefore sin r= (sin50^(@))/(1.55)=(0.7660)/(1.55)=0.4942` `therefore r=29^(@)37. andcosr = cos 29^(@)37.` `=0.8694` `sin(50^(@)-29^(@)37.)=sin(49^(@)60.-29^(@)37.)` `=sin(20^(@)(20^(@)23.))=0.3493` `therefore` using the VALUES of cos r and `sin(i-r)` in (1) we get `L.S=(0.10xx0.3493)/(0.8694)=0.040m` Lateral shift produced is 0.040m |
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