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Geometrical Optics question

2022 · 28 Jun · Shift 1 · Q54
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  5. /2022 · 28 Jun · Shift 1 · Q54

Geometrical Optics question

2022 · 28 Jun · Shift 1 · Q54

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
The refracting angle of a prism is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be -
  1. A
    180 −-− 2A
  2. B
    90 −-− A
  3. C
    180 + 2A
  4. D
    180 −-− 3A
View written solutionFree

Correct answer: A

  1. Use the prism formula at minimum deviation

For a prism of refracting angle AAA, at minimum deviation,

rac{A+\delta_m}{2}=i

and the refractive index is

μ=sin⁡isin⁡(A/2)=sin⁡(A+δm2)sin⁡(A/2).\mu=\frac{\sin i}{\sin(A/2)}=\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin(A/2)}.μ=sin(A/2)sini​=sin(A/2)sin(2A+δm​​)​.

Given,

μ=cot⁡(A2)=cos⁡(A/2)sin⁡(A/2).\mu=\cot\left(\frac A2\right)=\frac{\cos(A/2)}{\sin(A/2)}.μ=cot(2A​)=sin(A/2)cos(A/2)​.

So,

sin⁡(A+δm2)sin⁡(A/2)=cos⁡(A/2)sin⁡(A/2).\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin(A/2)}=\frac{\cos(A/2)}{\sin(A/2)}.sin(A/2)sin(2A+δm​​)​=sin(A/2)cos(A/2)​.

Hence,

sin⁡(A+δm2)=cos⁡(A2).\sin\left(\frac{A+\delta_m}{2}\right)=\cos\left(\frac A2\right).sin(2A+δm​​)=cos(2A​).
  1. Convert cosine to sine

Using

cos⁡(A2)=sin⁡(90∘−A2),\cos\left(\frac A2\right)=\sin\left(90^\circ-\frac A2\right),cos(2A​)=sin(90∘−2A​),

we get

sin⁡(A+δm2)=sin⁡(90∘−A2).\sin\left(\frac{A+\delta_m}{2}\right)=\sin\left(90^\circ-\frac A2\right).sin(2A+δm​​)=sin(90∘−2A​).

Taking the principal physical solution,

A+δm2=90∘−A2.\frac{A+\delta_m}{2}=90^\circ-\frac A2.2A+δm​​=90∘−2A​.

So,

A+δm=180∘−AA+\delta_m=180^\circ-AA+δm​=180∘−A δm=180∘−2A.\delta_m=180^\circ-2A.δm​=180∘−2A.
  1. Match with options

Thus the angle of minimum deviation is

180∘−2A\boxed{180^\circ-2A}180∘−2A​

which corresponds to Option A.

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