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A-glass-prism-made-from-a-material-of-refractive-index-1-55-has-a-refracting-angle-of-60-The-prism-is-immersed-in-water-of-refractive-index-1-33-Determine-the-angle-of-minimum-deviation-for-a-paralle

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Question Number 38923 by NECx last updated on 01/Jul/18
A glass prism made from a material of refractive index 1.55 has a refracting angle of 60°.The prism is immersed in water of refractive index 1.33.Determine the angle of minimum deviation for a parallel beam of light passing through the prism.
$${A}\:{glass}\:{prism}\:{made}\:{from}\:{a}\:{material} \\ $$$${of}\:{refractive}\:{index}\:\mathrm{1}.\mathrm{55}\:{has}\:{a} \\ $$$${refracting}\:{angle}\:{of}\:\mathrm{60}°.{The}\:{prism} \\ $$$${is}\:{immersed}\:{in}\:{water}\:{of}\:{refractive} \\ $$$${index}\:\mathrm{1}.\mathrm{33}.{Determine}\:{the}\:{angle}\:{of} \\ $$$${minimum}\:{deviation}\:{for}\:{a}\:{parallel} \\ $$$${beam}\:{of}\:{light}\:{passing}\:{through}\:{the} \\ $$$${prism}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18
μ=((sin(((A+δ_m )/2)))/(sin((A/2)))) ((1.55)/(1.33))=((sin(((60^o +δ_m )/2)))/(1/2)) sin(((60^o +δ_m )/2))=((1.55)/(2×1.33))=0.58≈sin36^o δ_m =12^o μ_w ×sini_1 =μ_g ×sinr_1 δ=i_1 +i_2 −A r_1 ^ =r_2 =(A/2) when deviation minimum i_1 =i_2 =((δ_m +A)/2) μ_w sin(((δ_m +A)/2))=μ_g sin((A/2)) 1.33×sin(((δ_m +60^o )/2))=1.55sin(((60^o )/2)) sin(((δ_m +60^o )/2))=((1.55)/(2×1.33))≈sin36^o δ_m =12^o
$$\mu=\frac{{sin}\left(\frac{{A}+\delta_{{m}} }{\mathrm{2}}\right)}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{1}.\mathrm{55}}{\mathrm{1}.\mathrm{33}}=\frac{{sin}\left(\frac{\mathrm{60}^{{o}} +\delta_{{m}} }{\mathrm{2}}\right)}{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${sin}\left(\frac{\mathrm{60}^{{o}} +\delta_{{m}} }{\mathrm{2}}\right)=\frac{\mathrm{1}.\mathrm{55}}{\mathrm{2}×\mathrm{1}.\mathrm{33}}=\mathrm{0}.\mathrm{58}\approx{sin}\mathrm{36}^{{o}} \\ $$$$\delta_{{m}} =\mathrm{12}^{{o}} \\ $$$$\mu_{{w}} ×{sini}_{\mathrm{1}} =\mu_{{g}} ×{sinr}_{\mathrm{1}} \\ $$$$\delta={i}_{\mathrm{1}} +{i}_{\mathrm{2}} −{A} \\ $$$$\overset{} {{r}}_{\mathrm{1}} ={r}_{\mathrm{2}} =\frac{{A}}{\mathrm{2}}\:\:\:{when}\:{deviation}\:{minimum} \\ $$$${i}_{\mathrm{1}} ={i}_{\mathrm{2}} =\frac{\delta_{{m}} +{A}}{\mathrm{2}} \\ $$$$\mu_{{w}} {sin}\left(\frac{\delta_{{m}} +{A}}{\mathrm{2}}\right)=\mu_{{g}} {sin}\left(\frac{{A}}{\mathrm{2}}\right) \\ $$$$\mathrm{1}.\mathrm{33}×{sin}\left(\frac{\delta_{{m}} +\mathrm{60}^{{o}} }{\mathrm{2}}\right)=\mathrm{1}.\mathrm{55}{sin}\left(\frac{\mathrm{60}^{{o}} }{\mathrm{2}}\right) \\ $$$${sin}\left(\frac{\delta_{{m}} +\mathrm{60}^{{o}} }{\mathrm{2}}\right)=\frac{\mathrm{1}.\mathrm{55}}{\mathrm{2}×\mathrm{1}.\mathrm{33}}\approx{sin}\mathrm{36}^{{o}} \\ $$$$\delta_{{m}} =\mathrm{12}^{{o}} \\ $$$$ \\ $$
Commented by NECx last updated on 04/Jul/18
wow....Thank you sir
$${wow}….{Thank}\:{you}\:{sir} \\ $$

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