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Question

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 3.00×108 ms1.


A
2.12×108 ms1
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B
2.50×108 ms1
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C
3.00×108 ms1
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D
3.50×108 ms1
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Solution

The correct option is A 2.12×108 ms1
Draw ray diagram of a light ray passing through the prism



From the triangle, A+(90r1)++(90r2)=180
r1+r2=A
And angle of deviation δ=δ1+δ2
=(ir1)+(er2)=i+eA
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+eA=2iA=δmini=A+δmin2
r1+r2=r+r=2r=Ar=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+302)sin(602)=2
We know μP=cvP
Therefore, velocity of light in prism vP=cμP=32×108
=2.12×108 m/s

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