Question Number 38923 by NECx last updated on 01/Jul/18
$${A}\:{glass}\:{prism}\:{made}\:{from}\:{a}\:{material} \\ $$$${of}\:{refractive}\:{index}\:\mathrm{1}.\mathrm{55}\:{has}\:{a} \\ $$$${refracting}\:{angle}\:{of}\:\mathrm{60}°.{The}\:{prism} \\ $$$${is}\:{immersed}\:{in}\:{water}\:{of}\:{refractive} \\ $$$${index}\:\mathrm{1}.\mathrm{33}.{Determine}\:{the}\:{angle}\:{of} \\ $$$${minimum}\:{deviation}\:{for}\:{a}\:{parallel} \\ $$$${beam}\:{of}\:{light}\:{passing}\:{through}\:{the} \\ $$$${prism}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18
$$\mu=\frac{{sin}\left(\frac{{A}+\delta_{{m}} }{\mathrm{2}}\right)}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{1}.\mathrm{55}}{\mathrm{1}.\mathrm{33}}=\frac{{sin}\left(\frac{\mathrm{60}^{{o}} +\delta_{{m}} }{\mathrm{2}}\right)}{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${sin}\left(\frac{\mathrm{60}^{{o}} +\delta_{{m}} }{\mathrm{2}}\right)=\frac{\mathrm{1}.\mathrm{55}}{\mathrm{2}×\mathrm{1}.\mathrm{33}}=\mathrm{0}.\mathrm{58}\approx{sin}\mathrm{36}^{{o}} \\ $$$$\delta_{{m}} =\mathrm{12}^{{o}} \\ $$$$\mu_{{w}} ×{sini}_{\mathrm{1}} =\mu_{{g}} ×{sinr}_{\mathrm{1}} \\ $$$$\delta={i}_{\mathrm{1}} +{i}_{\mathrm{2}} −{A} \\ $$$$\overset{} {{r}}_{\mathrm{1}} ={r}_{\mathrm{2}} =\frac{{A}}{\mathrm{2}}\:\:\:{when}\:{deviation}\:{minimum} \\ $$$${i}_{\mathrm{1}} ={i}_{\mathrm{2}} =\frac{\delta_{{m}} +{A}}{\mathrm{2}} \\ $$$$\mu_{{w}} {sin}\left(\frac{\delta_{{m}} +{A}}{\mathrm{2}}\right)=\mu_{{g}} {sin}\left(\frac{{A}}{\mathrm{2}}\right) \\ $$$$\mathrm{1}.\mathrm{33}×{sin}\left(\frac{\delta_{{m}} +\mathrm{60}^{{o}} }{\mathrm{2}}\right)=\mathrm{1}.\mathrm{55}{sin}\left(\frac{\mathrm{60}^{{o}} }{\mathrm{2}}\right) \\ $$$${sin}\left(\frac{\delta_{{m}} +\mathrm{60}^{{o}} }{\mathrm{2}}\right)=\frac{\mathrm{1}.\mathrm{55}}{\mathrm{2}×\mathrm{1}.\mathrm{33}}\approx{sin}\mathrm{36}^{{o}} \\ $$$$\delta_{{m}} =\mathrm{12}^{{o}} \\ $$$$ \\ $$
Commented by NECx last updated on 04/Jul/18
$${wow}....{Thank}\:{you}\:{sir} \\ $$