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Question: The speed of light in air is \(3 \times {10^8}\)m/s. If the same light undergoes minimum deviation b...

The speed of light in air is 3×1083 \times {10^8}m/s. If the same light undergoes minimum deviation by 60 degree in a an equilateral glass prism, its speed in prism will be

  1. 3×1083 \times {10^8}m/s
  2. 3×108\sqrt 3 \times {10^8}m/s
  3. 13×108\dfrac{1}{{\sqrt 3 }} \times {10^8}m/s
  4. 33×1083\sqrt 3 \times {10^8}m/s
Explanation

Solution

Use the formula of minimum deviation and find out the value of μ  \mu \;. Then use the relation μ  =cv\mu \; = \dfrac{c}{v} to find out the value of V.

Complete step-by-step answer:
Formula required:
μ  =sin⁡(A+∂2)sin⁡(A2)\mu \; = \dfrac{{\sin (\dfrac{{A + \partial }}{2})}}{{\sin (\dfrac{A}{2})}} to find out the refractive index.
μ  =cv\mu \; = \dfrac{c}{v} to find out the value of v.
Given that the angle of prism =60∘{60^ \circ }
Also given that the angle of minimum deviation = 60∘{60^ \circ }
Speed of light = 3×1083 \times {10^8}m/s
To find out the refractive index of the medium we use the reqd, formula :
μ  =sin⁡(A+∂2)sin⁡(A2)\mu \; = \dfrac{{\sin (\dfrac{{A + \partial }}{2})}}{{\sin (\dfrac{A}{2})}}
Putting the values in the above eqn, we have:
⇒μ  =sin⁡(60+602)sin⁡(602)\Rightarrow \mu \; = \dfrac{{\sin (\dfrac{{60 + 60}}{2})}}{{\sin (\dfrac{{60}}{2})}}
Further simplifying we get :

⇒μ  =sin⁡(60)sin⁡(30) ⇒μ  =3212 ⇒μ  =3  \Rightarrow \mu \; = \dfrac{{\sin (60)}}{{\sin (30)}} \\\ \Rightarrow \mu \; = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}} \\\ \Rightarrow \mu \; = \sqrt 3 \\\

Hence the refractive index of the prism = μ  =3\mu \; = \sqrt 3
Now finding out the velocity or speed of light in the given medium by the mentioned formula we have,
μ  =cv\mu \; = \dfrac{c}{v}
⇒3=3×108v ⇒v=3×108ms−1.  \Rightarrow \sqrt 3 = \dfrac{{3 \times {{10}^8}}}{v} \\\ \Rightarrow v = \sqrt 3 \times {10^8}m{s^{ - 1}}. \\\
Hence the refractive index of the prism = μ  =3\mu \; = \sqrt 3 and the velocity is 3×108\sqrt 3 \times {10^8}m/s.

Hence the correct answer is option 2) 3×108\sqrt 3 \times {10^8}m/s.

Note:
If the prism is thin (prism angles up to 5∘{5^ \circ }) then we can use the direct formula to find refractive index by using the formula ∂m=A(μ  −1){\partial _m} = A(\mu \; - 1).
This formula is a special case of prism which is derived from the formula μ  =sin⁡(A+∂2)sin⁡(A2)\mu \; = \dfrac{{\sin (\dfrac{{A + \partial }}{2})}}{{\sin (\dfrac{A}{2})}}. Since the prism angle is very small, hence just substitute sin A = A and solve the equation and find out the value of the refractive index.