A parallel beam of light is incident from air at an angle - on the side PQ of a right angles triangular prism of refractive index n - -2- Light undergoes total internal reflection in the prism at the face PR when - has a minimum value of 45o- The angle - of the prism is

The correct option is A 15^o From the figure, we can see #160; sin ^ 1 1 √( 2 ) #160; #160; = 45 ^∘ #160; Using geometry to find angle ...

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Standard X

Physics

Critical Angle

A parallel be...

Question

A parallel beam of light is incident from air at an angle $α$ on the side $P Q$ of a right angles triangular prism of refractive index $n = \sqrt{2} .$ Light undergoes total internal reflection in the prism at the face $P R$ when $α$ has a minimum value of $45^{o} .$ The angle $θ$ of the prism is

15^{o}

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{22.5}^{o}

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30^{o}

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45^{o}

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Solution

The correct option is A $15^{o}$
From the figure, we can see
${sin}^{- 1} \frac{- 1}{\sqrt{2}} = 45^{\circ}$
Using geometry to find angle $A B C$ .
$= 60 - θ - (i)$
and i.e., $= 45^{\circ} - (i i)$
Equating $(i)$ and $(i i)$
$45 = 60 - θ θ = 15^{\circ}$

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A parallel beam of light is incident from air at an angle α on the side PQ of a right angles triangular prism of refractive index n=√2. Light undergoes total internal reflection in the prism at the face PR when α has a minimum value of 45o. The angle θ of the prism is

The correct option is A 15oFrom the figure, we can see sin−1−1√2=45∘Using geometry to find angle ABC.=60−θ−(i)and i.e.,=45∘−(ii)Equating (i) and (ii)45=60−θθ=15∘

The correct option is A $15^{o}$
From the figure, we can see
${sin}^{- 1} \frac{- 1}{\sqrt{2}} = 45^{\circ}$
Using geometry to find angle $A B C$ .
$= 60 - θ - (i)$
and i.e., $= 45^{\circ} - (i i)$
Equating $(i)$ and $(i i)$
$45 = 60 - θ θ = 15^{\circ}$