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The thickness of a glass slab is 0.25 m. It has a refractive index of 1.5 A ray of light is incident on the surface of the slab at an angle of 60^(@). Find the lateral displacement of the light when it emerges from the other side of the mirror. |
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Answer» Solution :Give: thickness of the lab, (t) =0.25 m, REFRACTIVE index, (n) =1.5 angle of INCIDENCE (i) = `60^(@)`. Using Snell.s law, `1 xx sin I = n sin R` `sin r = (sin r)/(n) = (sin 60)/(1.5) = 0.58` `r = sin^(-1)(0.58) = 35,256(@)` Latereal displacement is, `L=t((sin(i-r))/(COS(r)))` `L=(0.25)xx((sin(60-35.25))/(cos(35.25)))=0.1281m` The lateral displacement is, `L = 12.81 cm` |
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