Dispersive Power 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.
Dispersive Power 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 Dispersive Power |
| Academic Session | 2025-2026 |
Numericals on Dispersive Power
Class-12 Nootan ISC Physics Solution Ch-17 Refraction and Dispersion of Light through a Prism
Que-18: Calculate dispersive power of a transparent material. (given : nv = 1.56 , nr = 1.54 , ny = 1.55) (ISC 2016)
Ans- w = nv – nr / ny – 1
=> 1.56-1.54 / 1.55-1
=> 0.036
Que-19: Calculate dispersive power of a transparent material whose refractive indices for violet and red colours are 1.58 and 1.56 respectively. (ISC 2021)
Ans- w = nv – nr / nv +nr – 1
=> 1.58-1.56 / 1.58+1.56-1
=> 0.0093
Que-20: The deviation produced for violet, yellow and red lights for crown glass are 3.75°, 3.25° and 2.86°, respectively. Calculate the dispersive power of the crown glass. (ISC 2019)
Ans- w = δv-δr / δy-1
=> 3.75-2.86 / 3.25-1
=> 0.274
Que-21: The refractive indices of a material for red, violet and yellow lights are 1.52, 1.62 and 1.59, respectively. Calculate the dispersive power of the material. If the mean deviation is 40°, then what will be the angular dispersion produced by a prism of this material?
Ans- w = 1.62-1.52 / 1.59-1
=> 0.169
angular dispersion
θ = w . δy
=> 0.169 x 40
=> 6.76°
— : End of Dispersive Power Numerical Class-12 Nootan ISC Physics Solution Ch-17 Refraction and Dispersion of Light through a Prism:–
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