Interference of Light Miscellaneous Numerical Class-12 Nootan ISC Physics Solution Ch-20. 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.
Interference of Light Miscellaneous Numerical Class-12 Nootan ISC Physics Solution Ch-20
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-20 | Interference of Light |
| Topics | Miscellaneous Numericals |
| Academic Session | 2025-2026 |
Miscellaneous Numericals
Interference of Light Class-12 Nootan ISC Physics Solution Ch-20
Que-18: In Young’s double-slit experiment the distance between the centres of the adjacent fringes is 0.10 mm. If the distance of the screen from the slits is doubled, the distance between the slits is halved and the wavelength of light changed from 6.4 x 10^-7 m to 4.0 x 10^-7 m, then what is the new distance between the fringes?
Ans- β1/β2 = D1/D2 x λ1/λ2 x d2/d1
=> 10/β2 = D1/2D1 x (6.4 x 10^-7 / 4 x 10^-7) x d1/2d1
=> β2 = 0.25 mm
Que-19: In Young’s double-slit experiment, the slits are 1.5 mm apart and the screen is placed 1 m away from the slits. When the slits are illuminated by monochromatic light, the width of 10 fringes on the screen is measured as 3.93 mm. Find the wavelength of light used. What will be the width of 10 fringes when the screen is placed 1.5 m away from the slits and the same light is used?
Ans- β = 3.93 x 10^-3 / 10 = Dλ/d = 1 x λ / 1.5 x 10^-3
=> λ = 5895 Å
β1/β2 = D1/D2 => 10β1/10β2 = D1/D2
3.93/10β2 = 1/1.5 => 10β2 = 5.895 mm
Que-20: In an experiment with sodium light (5890 Å), an interference pattern is obtained in which 20 equally spaced fringes occupy 2.30 cm on the screen. On replacing sodium light with another monochromatic source, 30 fringes occupy 2.80 cm on the screen. Find the wavelength of the light from the second source.
Ans- β1 = 2.3/20 , β2 = 2.8/30
now λ1/λ2 = β1/β2 => 5890Å / λ2 = 2.3/20 x 30/2.8
=> λ2 = 4780 Å
Ans21: A double-slit is illuminated by sodium light (λ = 5893 Å). In the double-slit the distance between the two slits is 0.050 cm. The interference pattern is obtained on a screen placed at a distance of 1.00 m. Calculate the distance between the fifth bright fringe on one side of the central fringe and the fifth bright fringe on the other side.
Ans- distance between the fifth bright fringe on one side of the central fringe and the fifth bright fringe on the other side
=> x = 2 * nDλ/d
= 2 * 5 * 1 * 5893 * 10^-10 / 0.050 * 10^-2
= 1.18 cm
Que-22: In Young’s double-slit experiment, the slits are 0.05 cm apart and the interference fringes are obtained on a screen 1 m away from the slits. The slits are illuminated by sodium light (5893 Å). Find the distance between 4th bright fringe on one side and 3rd bright fringe on the other side of the central bright fringe.
Ans- Distance = 3Dλ/d + 4Dλ/d
= 7Dλ/d = 7 x 1 x 5893 x 10^-10 / 0.05 x 10^-2
= 8.25 mm
Que-23: In Young’s double-slit experiment, the separation between the slits is 0.15 mm and the fringes are obtained on a screen distant 0.75 m from the slits. If the third dark fringe is at a distance of 5.5 mm from the central (bright) fringe, find the wavelength of light used. What will be the distance of the same dark fringe from the centre, if the entire apparatus is immersed in water? The refractive index of water is 4/3.
Ans- x = (n-1/2) Dλ/d
=> λ = xd/(n-1/2)D = 5.5 * 10^-3 * 0.15 * 10^-3 / 2.5 * 0.75
= 4400 Å
when apparatus dipped in water then,
λ’=λ/μ = 4400/(4/3) = 3300 Å
∴ x = 2.5 * Dλ / d = 2.5 * 0.75 * 3300 * 10^-10 / 0.15 * 10^-3
= 4.125 mm
–: Interference of Light Miscellaneous Numerical Class-12 Nootan ISC Physics Solution Ch-20. :–
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