CBSE Class 12 Physics 2019 Outside Delhi Set 2 Paper

Question : 8 of 9

Marks: +1, -0

A ray of light incident on the face

A B

of an isosceles triangular prism makes an angle of incidence (i) and deviates by angle

β

as shown in the figure. Show that in the position of minimum deviation

∠ β=∠ α

. Also find out the condition when the refracted ray

Q R

suffers total internal reflection.

Solution:

Proving

α=β

Finding

i_c

For minimum deviation,

\;r_1+r_2=A ; \;\; r_1=r_2

\; (90^{∘}-β) + (90^{∘}-β) =A

\;180^{∘}-2 β=A

\;2 β=180^{∘}-A

\;2 β=2 α

\;β=α

\;r_1+r_2=A

\;r_1+i_c=A

\;i_c=A-r_1

\;i_c=A- (90^{∘}-β)

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