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Geometrical Optics question

2024 · 31 Jan · Shift 1 · Q70
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Geometrical Optics question

2024 · 31 Jan · Shift 1 · Q70

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
The refractive index of a prism with apex angle AAA is cot⁡A/2\cot A / 2cotA/2. The angle of minimum deviation is :
  1. A
    δm=180∘−3 A\delta_m=180^{\circ}-3 \mathrm{~A}δm​=180∘−3 A
  2. B
    δm=180∘−4A\delta_m=180^{\circ}-4 Aδm​=180∘−4A
  3. C
    δm=180∘−2A\delta_m=180^{\circ}-2 Aδm​=180∘−2A
  4. D
    δm=180∘−A\delta_m=180^{\circ}-Aδm​=180∘−A
View written solutionFree

Correct answer: C

  1. Use the prism formula at minimum deviation

For a prism in air, the refractive index is

μ=sin⁡(A+δm2)sin⁡(A2)\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}μ=sin(2A​)sin(2A+δm​​)​

Given:

μ=cot⁡A2\mu = \cot \frac{A}{2}μ=cot2A​

So,

sin⁡(A+δm2)sin⁡(A2)=cot⁡A2\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}=\cot\frac{A}{2}sin(2A​)sin(2A+δm​​)​=cot2A​
  1. Substitute cot⁡A2\cot\frac{A}{2}cot2A​

Recall,

cot⁡A2=cos⁡A2sin⁡A2\cot\frac{A}{2} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}cot2A​=sin2A​cos2A​​

Thus,

sin⁡(A+δm2)sin⁡(A2)=cos⁡A2sin⁡A2\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}sin(2A​)sin(2A+δm​​)​=sin2A​cos2A​​

Multiplying both sides by sin⁡A2\sin\frac{A}{2}sin2A​,

sin⁡(A+δm2)=cos⁡A2\sin\left(\frac{A+\delta_m}{2}\right)=\cos\frac{A}{2}sin(2A+δm​​)=cos2A​
  1. Convert cosine to sine

Using

cos⁡θ=sin⁡(90∘−θ)\cos\theta = \sin(90^\circ-\theta)cosθ=sin(90∘−θ)

we get

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

For the physical prism case, take the principal solution:

A+δm2=90∘−A2\frac{A+\delta_m}{2}=90^\circ-\frac{A}{2}2A+δm​​=90∘−2A​
  1. Solve for δm\delta_mδm​
A+δm=180∘−AA+\delta_m = 180^\circ - AA+δm​=180∘−A δm=180∘−2A\delta_m = 180^\circ - 2Aδm​=180∘−2A
  1. Match with options

This corresponds to:

δm=180∘−2A\boxed{\delta_m = 180^\circ - 2A}δm​=180∘−2A​

So the correct option is C.

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