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A ray of light is incident at an angle of `60^(@)` on one face of a rectangular glass slab of thickness `0.1 m`, and refractive index `1.5`.Calculate the lateral shift produced. |
Answer» Correct Answer - `0.0513 m` Here, `i_(1) = 60^(@), t = 0.1m, mu = 1.5` As `(sin i_(1))/(sin r_(1)) = mu` `:. Sin r_(1) = (sin i_(1))/(mu) = (sin 60^(@))/(1.5) = 0.5773` `r_(1) = sin^(-1)(0.5773) = 35.3^(@)` Lateral shift `= (t sin (t_(1) - r_(1)))/(cos r_(1))` `= (0.1 sin (60^(@) - 35.3^(@)))/(cos 35.3^(@)) = (0.1 sin 24.7^(@))/(cos 35.3^(@))` `= (0.1 xx 0.418)/(0.816) = 0.0513 m` |
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