Superposition of Waves Numerical Class-12 Nootan ISC Physics Solution Ch-20 Interference of Light. 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.
Superposition of Waves Numerical Class-12 Nootan ISC Physics Solution Ch-20 Interference of Light
| Board | 12 |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-20 | Interference of Light |
| Topics | Numericals on Superposition of Waves |
| Academic Session | 2025-2026 |
Numericals on Superposition of Waves
Class-12 Nootan ISC Physics Solution Ch-20 Interference of Light.
Que-1: Two waves of equal frequencies have their amplitudes in the ratio of 3: 5. They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.
Ans- a1/a2 = 3/5
=> Imax / Imin = (a1+a2 / a1-a2)² = (8/2)² = 16:1
Que-2: Two coherent light sources whose intensity ratio is 81: 1 produce interference fringes. Calculate the ratio of intensities of maxima and minima in the fringe-system.
Ans- I1 / I2 = 81/1 => √I1 / √I2 = 9/1
Imax / I min = (√I1+√I2 / √I1-√I2)² = (9+1 / 9-1)² = 25:16
Que-3: The two slits in Young’s double-slit experiment have widths in the ratio 9 : 4. Find the ratio of light intensities at maxima and minima in the interference pattern.
Ans- (a1)²/(a2)² = 9/4 => a1/a2 = 3/2
Imax / Imin = (a1+a2 / a1-a2)² = (3+2 / 3-2)² = 25:1
Que-4: The intensity ratio of maxima and minima in an interference pattern is 4: 1. Find the ratio of slit widths.
Ans- Imax / Imin = 4/1 => √Imax / √Imin = 2/1
∴ a1/a2 = √Imax +√Imin / √Imax -√Imin = 3/1
∴ratio of slit width = (a1/a2)² = 9:1
–: End of Superposition of Waves Numerical Class-12 Nootan ISC Physics Solution Ch-20 Interference of Light :—
Return to : – Nootan Solutions for ISC Class-12 Physics
Thanks
Please share with your friends
