Formula for Refractive Index of Prism Numerical Class-12 Nootan ISC Physics Solution Ch-17 Refraction and Dispersion of Light through a Prism. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Formula for Refractive Index of Prism Numerical Class-12 Nootan ISC Physics Solution Ch-17 Refraction and Dispersion of Light through a Prism
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-17 | Refraction and Dispersion of Light through a Prism |
| Topics | Numerical on Formula for Refractive Index of Prism |
| Academic Session | 2025-2026 |
Numerical on Formula for Refractive Index of Prism
Class-12 Nootan ISC Physics Solution Ch-17 Refraction and Dispersion of Light through a Prism
Que-3: The angle of a prism is 60° and the angle of minimum deviation is 39°. What is the refractive index of the material of the prism? (sin 49.5° = 0.76)
Ans- μ = [sin (A+δm)/2] / sin (A/2)
=> sin (60+39/2) / sin 30
=> sin 49.5 / sin 30
=> 0.76 x 2 = 1.52
Que-4: Calculate the refractive index of the material of an equilateral prism for which the angle of minimum deviation is 30°. (ISC 2022)
Ans- μ = [sin (A+δm)/2] / sin (A/2)
=> sin (60+30/2) / sin 30
=> sin 45 / sin 30
=> √2
Que-5: A certain equilateral prism of glass has a refractive index of 1.60. Calculate the angle of minimum deviation produced by it. (ISC 2021)
Ans- μ = [sin (A+δm)/2] / sin (A/2)
=> 1.60 = sin (60°+δm/2) / sin 30°
=> 1.60 = sin (60°+δm/2) / 1/2
=> sin (60°+δm/2) = 0.8
=> 60°+δm/2 = 53.13°
now 60°+δm/2 = 2 x 53.13
=> δm = 46.3°
Que-6: Calculate the angle of minimum deviation for a glass prism of angle 60°, the refractive index of glass being 1.658. What is the corresponding angle of incidence? Given sin 56° = 0.829.
Ans- μ = [sin (A+δm)/2] / sin (A/2)
=> sin (A+δm)/2 / μ sin (A/2)
=> 1.658 x 1/2 = 0.829 = sin 56
=> A+δm = 56 x 2 = 112
=> δm = 112 – 60 = 25°
again 2i = A+δm
=> 2i = 60+52
=> i = 56°
Que-7: A regular 60° glass prism has refractive index √2 for a given wavelength. Calculate the angle of minimum deviation δm. At what angle of incidence should a monochromatic light of this wavelength fall on the face of the prism so as to get a minimum deviation? (use sin 30° = 1/2 and sin 45° = 1/√2)
Ans- μ = [sin (A+δm)/2] / sin (A/2)
=>√2= sin (60°+δm/2) / sin 30°
=> √2= sin (60°+δm/2) / 1/2
=> sin (60°+δm/2) = √2/2
=> 60°+δm/2 = 45°
now 60°+δm/2 = 2 x 45°
=> δm = 90° – 60° = 30°
again 2i = A+δm
=> 2i = 60+30
=> i = 45°
Que-8: What should be the refracting angle of a prism of refractive index 1.5 so that a light ray incident on the prism at an angle of 50° suffers minimum deviation?
Ans- μ = [sin (A+δm)/2]
i = e = A+δm/2
=> 50° = A+δm/2
=> δm = 100°-A
μ = [sin (A+δm)/2] / sin (A/2)
=> 1.5= sin (A+(100-A)/2) / sin (A/2)
=> sin (A/2) = sin 50° /1.5
=> sin (A/2)= 0.766/1.5
=> A = 61.4°
Que-9: The angle of a glass prism is 59.5°. The angle of minimum deviation is 38.5°. Without using the value of refractive index of the glass, calculate the angle of incidence for a ray of light suffering minimum deviation. (ISC 2007)
Ans- i = A+δm/2
=> 59.5+38.5 / 2
=> 98/2 = 49°
Que-10: A ray of light incident at 50° on a 60° prism suffers minimum deviation. Find the angle of minimum deviation and the refractive index of the material of the prism.
Ans- i = A+δm/2
=> δm = 2i-A
=> 2 x 50 -60 = 40°
again μ = sin i / sin r = sin 50 / sin 30
=> 0.766 x 2 = 1.532
Que-11: A ray of light falls on an equilateral prism in the position of minimum deviation. If the angle of incidence is 3/4 times angle of prism, then calculate angle of minimum deviation and refractive index of the material of the prism.
Ans- i = A+δm/2
=> 45° = 60° + δm / 2
=> 90° = 60° + δm
=> δm = 90° – 60° = 30°
μ = [sin (A+δm)/2] / sin (A/2)
=> sin (60°+30°/2) / sin 30°
=> sin 45° / sin 30°
=> (1/√2) / (1/2) = √2 = 1.41
Que-12: Refracting angle of a prism is A and refractive index of material of prism is 2 cos (A/2). Find the angle of minimum deviation for prism in terms of prism angle A.
Ans- μ = [sin (A+δm)/2] / sin (A/2)
=> 2 cos(A/2) = sin (A+δm)/2] / sin (A/2)
=> sin (A+δm)/2] = 2 sin (A/2) cos (A/2)
=> sin (A+δm)/2] / sin (A)
=> (A+δm)/2 = 2 A
=> δm = 2A-A
=> δm = A
Que-13: Find out the relation between the refractive index (n) of the glass prism and prism angle A for the case, when the angle of prism A is equal to the angle of minimum deviation δm. Hence, obtain the value of the refractive index if prism is an equilateral prism.
Ans- For an equilateral prism, A = 60°
=> n = [sin (A+δm)/2] / sin (A/2)
=> n = sin (A+A/2] / sin (A/2)
=> n = sin (A) / sin (A/2)
=> n = 2 sin (A/2) cos (A/2) / sin (A/2)
=> n = 2 cos (A/2)
=> n = 2 x √3/2 = √3 = 1.732
— : End Formula for Refractive Index of Prism Numerical Class-12 Nootan ISC Physics Solution Ch-17:–
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