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Question

- 50 ∘
- 90∘
- 60∘
- 40∘

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Solution

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sin 60osin r1=√3

sin r1=12⟹r1=30o

As we know that r1=r2⟹r2=30o

Using Snell's Law at point Q we get:-

sin 30osin i2=1√3⟹i2=60o

Since reflection at point Q occurs

⟹r′2=r2=60o

Let angle between the refracted ray and reflected ray be α

α=180o−(r′2+r2)=90o

Hence option (B) is correct.

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