Refraction through 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.
Refraction through 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 | Numericals on Refraction through Prism |
| Academic Session | 2025-2026 |
Numericals on Refraction through Prism
Class-12 Nootan ISC Physics Solution Ch-17 Refraction and Dispersion of Light through a Prism
Que-1: A ray of light is incident on a 60° prism of glass of refractive index 1.5 at an angle of 40°. Find the angle of emergence of the ray.
Ans- n = sin i/ sin r => sin r = sin i / n
=> sin 40/1.5 = 0.4285
r= 25.4°
A = r + r’
=> r’ = A-r = 60-25.4 = 34.6°
n = sin i’/ sin r’
=> sin i’ = n × sin r’
=> 1.5× sin 34.6 = 0.85176
=> i’= 58.5 °
Que-2: The refracting angle of a glass prism of refractive index 1.45 is 60°. At what angle should a light ray be incident at the first face of the prism so that it emerges at the second face just parallel to the face?
Ans- Given , Refractive index = ng = 1.45 ,
Temperature = A= 60° , e = 90°
r2 = C (Critical angle)
=> sin C= 1/ng = sin r2 = 1/1.45 —– (i)
Also, sin i / sin r1 = ng = 1.45
Thus, sin i = 1.45 × sin r1
Now, r1+r2 = A = 60°
=> r1 = 60° −r2
Thus, sin i = 1.45 × sin (60°−r2)
Expanding the equation in (i),
i = sin^-1 [1.45 × sin 60 √1− 1/1.45²−cos 60°]
=> sin^-1(0.4) = 24.1°
— : End of Refraction through Prism Numerical Class-12 Nootan ISC Physics Solution Ch-17 :–
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