• # question_answer A glass prism of refracting angle $60{}^\circ$ and refractive index 1.5, is completely immersed in water of refractive index 1.33. Calculate the angle of minimum deviation of the prism in this situation. $(given,si{{n}^{-1}}0.56=34.3{}^\circ )$ Or (i) The refractive indices of crown glass for red and violet colours are 1.515 and 1.523 respectively. Find the dispersive power of crown glass between these colours. (ii) If a crown glass prism produces a mean deviation of $40{}^\circ ,$ what will be the angular dispersion?

We know that,    ${}^{w}{{\mu }_{g}}=\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$ $\frac{{}^{a}{{\mu }_{g}}}{{}^{a}{{\mu }_{w}}}=\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin (A/2)}$             $\Rightarrow$               $\frac{1.5}{1.33}=\frac{\sin \left( \frac{60{}^\circ +{{\delta }_{m}}}{2} \right)}{\sin (60{}^\circ /2)}$             or         $\sin \left( \frac{60{}^\circ +{{\delta }_{m}}}{2} \right)=\frac{1.5}{1.33}\times \frac{1}{2}=0.56$             or         $\left( \frac{A+{{\delta }_{m}}}{2} \right)={{\sin }^{-1}}(0.56)=34.3{}^\circ$             $\Rightarrow$ $A+{{\delta }_{m}}=68.6{}^\circ$ or ${{\delta }_{m}}=68.6{}^\circ -60{}^\circ =8.6{}^\circ$ (i) The dispersive power of crown glass is given by                         $\omega =\frac{{{\mu }_{V}}-{{\mu }_{R}}}{{{\mu }_{Y}}-1}$ Given,   ${{\mu }_{V}}=1.523;{{\mu }_{R}}=1.515$ So,       ${{\mu }_{Y}}=\frac{{{\mu }_{V}}+{{\mu }_{R}}}{2}=\frac{1.523+1.515}{2}$             = 1.519 $\therefore$      $\omega =\frac{1.523-1.515}{1.519-1}=\frac{0.008}{0.519}=0.0154$ (ii) If ${{\delta }_{V}},{{\delta }_{R}}$ and ${{\delta }_{Y}}$ be the deviations produced by the crown-glass prism in violet, red and yellow, respectively rays of light, then the dispersive power of crown glass between red and violet rays is             $\omega =\frac{{{\delta }_{V}}-{{\delta }_{R}}}{{{\delta }_{Y}}}$ $\therefore$ Angular dispersion between red and violet rays is             $\theta ={{\delta }_{V}}-{{\delta }_{R}}=\omega \times {{\delta }_{Y}}$             $=0.0154\times 40{}^\circ =0.616{}^\circ$
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