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A ray of light is incident on a prism at an angle 50^(@) and angle of prism is 60^(@)and RI 1.5. Calculate the angle of total deviation (for non symmetric refraction). |
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Answer» Solution :`A=60^(@), n=1.5, i_(1)=50^(@)` Applying Snell.slaw, `n=(sin i_(1))/(sin r_(1))` `sin r_(1)= (sin50^(@))/(1.5) = (0.7660)/(1.5) = 0.5106` `r_(1)= sin^(-1) 0.5106` `r_(1)=30^(@)42.` But, `A=r_(1)+r_(2)` `r_(2)=A-r_(1)=59^(@)60. - 30^(@)42.=29^(@)18.` Also,`n=(sin i_(2))/(sin r_(2))` `1.5=(sin i_(2))/(sin29^(@) 18.)` `sin i_(2)= 1.5xx0.4894=0.7341` `i_(2)=sin^(-1)(0.7341)=47^(@)14.` `THEREFORE d=i_(1)+i_(2)-A=50+47^(@)14. 60^(@)=37^(@)14.` |
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