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A ray of light is incident at an angle of `45^(@)` on one face of a rectangular glass slab of thickness `10 cm` and refractive index `1.5`. Calculate the lateral shift produced. |
Answer» Here, `i_(1) = 45^(@), t = 10 cm = 0.1 m` `mu = 1.5`, lateral shift `= ?` As `mu = (sin i_(1))/(sin r_(1))` `:. sin r_(1) = (sin i_(1))/(mu) = (sin 45^(@))/(1.5) = (0.707)/(1.5) = 0.4713` `r_(1) = sin^(-1) (0.4713) = 28.14^(@)` lateral shift `= (t sin (i_(1) - r_(1)))/(cos r_(1))` `= (0.1 sin (45^(@) - 28.14^(@)))/(cos 28.14^(@))` `= (0.1 sin 16.86^(@))/(cos 28.14^(@)) = (0.1 xx 0.2900)/(0.8818) = 0.033 m` |
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