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Question Number 38923 by NECx last updated on 01/Jul/18

$$Aglassprismmadefromamaterial$$$$ofrefractiveindex1.55hasa$$$$refractingangleof60°.Theprism$$$$isimmersedinwaterofrefractive$$$$index1.33.Determinetheangleof$$$$minimumdeviationforaparallel$$$$beamoflightpassingthroughthe$$$$prism.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

$$\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$$$$\frac{1.55}{1.33}=\frac{sin\left(\frac{{60}^o+{\delta }_m}{2}\right)}{\frac{1}{2}}$$$$sin\left(\frac{{60}^o+{\delta }_m}{2}\right)=\frac{1.55}{2\times 1.33}=0.58\approx sin{36}^o$$$${\delta }_m={12}^o$$$${\mu }_w\times {sini}_1={\mu }_g\times {sinr}_1$$$$\delta =i_1+i_2-A$$$${\stackrel{}{r}}_1=r_2=\frac{A}{2}whendeviationminimum$$$$i_1=i_2=\frac{{\delta }_m+A}{2}$$$${\mu }_{w}sin\left(\frac{{\delta }_m+A}{2}\right)={\mu }_{g}sin\left(\frac{A}{2}\right)$$$$1.33\times sin\left(\frac{{\delta }_m+{60}^o}{2}\right)=1.55sin\left(\frac{{60}^o}{2}\right)$$$$sin\left(\frac{{\delta }_m+{60}^o}{2}\right)=\frac{1.55}{2\times 1.33}\approx sin{36}^o$$$${\delta }_m={12}^o$$$$$$

Commented by NECx last updated on 04/Jul/18

$$wow....Thankyousir$$

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