A ray of light is incident an 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.

Explanation

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. and cosr = 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