# A ray of light incident on the face AB of an isosceles triangular prism makes an angle of incidence (i) and deviates by angle $\beta$ as shown in the figure. Show that in the position of minimum deviation $\angle \beta =\angle \alpha$ Also find out the condition when the refracted ray QR suffers total internal reflection.

From the given figure, we have

$180-A = 2\alpha$

Now, for the minimum deviation:-

We have,

$r_{1}+r_{2}=A\; \; \; \; \; (\because r_{1}=90-\beta )$

then, $(90-\beta )+(90-\beta )=A$

$180-2\beta =A$

$2\beta =180-A$

$2\beta =2\alpha \; \; \; (\because 180-A=2\alpha )$

$\beta =\alpha$

Hence, proved:-

To find the total internal reflection;

we have,

$r_{1}+r_{2}=A$

$r_{1}+i_{c}=A$

$i_{c}=A-r_{1}$

$i_{c}=A-(90-\beta )\; \; \; \; (\because r_{1}=r_{2}=90-\beta )$

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