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Geometrical Optics question

2024 · 27 Jan · Shift 1 · Q63
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  5. /2024 · 27 Jan · Shift 1 · Q63

Geometrical Optics question

2024 · 27 Jan · Shift 1 · Q63

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
If the refractive index of the material of a prism is cot⁡(A2)\cot \left(\frac{A}{2}\right)cot(2A​), where AAA is the angle of prism then the angle of minimum deviation will be
  1. A
    π−2 A\pi-2 \mathrm{~A}π−2 A
  2. B
    π2−2 A\frac{\pi}{2}-2 \mathrm{~A}2π​−2 A
  3. C
    π−A\pi-\mathrm{A}π−A
  4. D
    π2−A\frac{\pi}{2}-\mathrm{A}2π​−A
View written solutionFree

Correct answer: A

  1. Use the prism formula at minimum deviation

For a prism in air, at minimum deviation,

mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

where:

  • μ\muμ = refractive index of prism material
  • AAA = angle of prism
  • δm\delta_mδm​ = minimum deviation

Given:

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

So,

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

Since

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

we get

\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}= rac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}

Multiplying by sin⁡(A2)\sin\left(\frac{A}{2}\right)sin(2A​),

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

Now use

cos⁡(A2)=sin⁡(π2−A2)\cos\left(\frac{A}{2}\right)=\sin\left(\frac{\pi}{2}-\frac{A}{2}\right)cos(2A​)=sin(2π​−2A​)

so

sin⁡(A+δm2)=sin⁡(π2−A2)\sin\left(\frac{A+\delta_m}{2}\right)=\sin\left(\frac{\pi}{2}-\frac{A}{2}\right)sin(2A+δm​​)=sin(2π​−2A​)
  1. Equate the principal angles

Taking the physically relevant principal solution,

A+δm2=π2−A2\frac{A+\delta_m}{2} = \frac{\pi}{2}-\frac{A}{2}2A+δm​​=2π​−2A​

Therefore,

A+δm=π−AA+\delta_m = \pi - AA+δm​=π−A δm=π−2A\delta_m = \pi - 2Aδm​=π−2A
  1. Check options
  • A: π−2A\pi-2Aπ−2A  correct
  • B: π2−2A\frac{\pi}{2}-2A2π​−2A  incorrect
  • C: π−A\pi-Aπ−A  incorrect
  • D: π2−A\frac{\pi}{2}-A2π​−A  incorrect

Hence, the angle of minimum deviation is

π−2A\boxed{\pi-2A}π−2A​
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