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Q. A certain prism produces a minimum deviation of $42^{\circ}$. It produces a deviation of $45^{\circ}$ when the angle of incidence is either $43^{\circ}$ or $62^{\circ}$. The angle of incidence when the prism undergoes minimum deviation is

KEAMKEAM 2020

Solution:

$\delta=i+e-A $
$\Rightarrow 45=43+62-A$
Or $A=43+62-45=60^{\circ}$
$D_{\min }=2 i-A$
or $i=\frac{A+A}{2}=\frac{60+42}{2}=51^{\circ}$