Diffraction of Light at Single Slit Numerical Class-12 Nootan ISC Physics Solution Ch-21. 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.
Diffraction of Light at Single Slit Numerical Class-12 Nootan ISC Physics Solution Ch-21
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-21 | Diffraction of Light |
| Topics | Numericals on Diffraction of Light at Single Slit |
| Academic Session | 2025-2026 |
Numericals on Diffraction of Light at Single Slit
Class-12 Nootan ISC Physics Solution Ch-21 Diffraction of Light
Que-1: Fraunhofer diffraction from a single-slit of width 1.0 µm is observed with light of wavelength 500 nm. Calculate the half angular width of the central maximum.
Ans- sin θ = λ/e
= 500 x 10^-9 / 1 x 10^-6 = 1/2
=> θ = 30°
Que-2: Fraunhofer diffraction from a single-slit of width 1.24 x 10^-6 m is observed with light of wavelength 6200 Å. Calculate the angular width of the central maximum.
Ans- sin θ = λ/e
= 6200 x 10^-10 / 1.24 x 10^-6 = 1/2
θ = 30°
∴ width of central max
= 2θ = 60°
Que-3: A parallel beam of light of wavelength 600 nm falls normally on a narrow slit of width 0.3 mm. Calculate the angular separation between (a) the first minimum and the central maximum, (b) the first subsidiary maxima on the two sides of the central maximum.
Ans- (a) θ = λ/e (for θ << sin θ = θ)
=> θ = 600 x 10^-9 / 0.3 x 10^-3
= 2 x 10^-3 rad
∴ width of first subsidiary maxima
= 2θ x 3/2 = 3θ = 3x 2 x 10^-3
= 6 x 10^-3 rad
Que-4: A parallel beam of light of wavelength 6 x 10^-5 cm falls normally on a straight slit of width 0.2 mm. Find the total angular width of the central diffraction maximum and also its linear width as observed on a screen placed 2 metre away.
Ans- width of central maxima
= 2θ = 2 x 6 x 10^-5 x 10^-2 / 2 x 10^-3
= 6 x 10^-3 rad
again linear width
= Dθ = 2 x 6 x 10^-3
= 1.2 cm
Que-5: A Fraunhofer diffraction pattern due to a single-slit of width 0.2 mm is being obtained on a screen placed at a distance of 2 metre from the slit. The first minima lie at 5 mm on either side of the central maximum on the screen. Find the wavelength of light.
Ans- Linear width = Dθ
= 5 x 10^-3 = 2 x λ/e
=> λ = 5 x 10^-3 x e / 2
= 5 x 10^-3 x 0.2 x 10^-3 / 2
= 5000 Å
Que-6: In a single slit diffraction experiment first minimum for λ1 = 600 nm coincides with first maxima for wavelength λ2, calculate λ2.
Ans- First minimum condition for λ1
a sin θ = 1 x λ1
a sin θ = 600 nm
First maximum condition for λ2
a sin θ = (1+1/2) λ2 (m=1)
a sin θ = 3/2 λ2
Equating both conditions
600 = 3/2 λ2
Solving for λ2
λ2 = 2/3 x 600
= 400 nm
–: Diffraction of Light at Single Slit Numerical Class-12 Nootan ISC Physics Solution Ch-21 ;–
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