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.

 

 

 

Answers (1)

 

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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