The angle of minimum deviation for prism angle π3 is π6. Calculate the velocity in the material of the prism, if the velocity of light in vacuum is 3.00×108ms−1.
A
2.12×108ms−1
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B
2.50×108ms−1
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C
3.00×108ms−1
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D
3.50×108ms−1
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Solution
The correct option is A2.12×108ms−1 Draw ray diagram of a light ray passing through the prism
From the triangle, A+(90∘−r1)++(90∘−r2)=180∘ ⇒r1+r2=A
And angle of deviation δ=δ1+δ2 =(i−r1)+(e−r2)=i+e−A
Variation of angle of deviation (δ) with angle of incidence (i) will look like this
So, δ=δmin at i=e. Also at i=e, i+e−A=2i−A=δmin⇒i=A+δmin2 r1+r2=r+r=2r=A⇒r=A2
Using Snell’s law, μ=sinisinr=sin(A+δmin2)sin(A2)
Here, μ=μPμM=μP because the medium is air
So, μP=sin(60∘+30∘2)sin(60∘2)=√2
We know μP=cvP
Therefore, velocity of light in prism vP=cμP=3√2×108 =2.12×108m/s