Numerical on Diffraction

JEE Mains and Advanced Numerical

1. Problem: A beam of light of wavelength 600 nm is incident on a single slit of width 0.5 mm. Calculate the angular width of the central maximum in the diffraction pattern.

Solution:

• Given:

  • Wavelength, (\lambda) = 600 nm = (6 \times 10^{-7}) m
  • Slit width, (a) = 0.5 mm = (5 \times 10^{-4}) m

• The angular width of the central maximum is given by: $$\theta = \frac{2\lambda}{a}$$

• Substituting the values, we get: $$\theta = \frac{2 \times 6 \times 10^{-7} \text{ m} }{5 \times 10^{-4} \text{ m}}$$ $$\theta = 2.4 \times 10^{-3}\text{ radians}$$ $$\theta = 1.37 \degree$$

Therefore, the angular width of the central maximum is (1.37\degree).

2. Problem: A parallel beam of light of wavelength 500 nm falls on a circular aperture of diameter 0.02 mm. Calculate the radius of the first dark ring in the diffraction pattern.

Solution:

• Given:

  • Wavelength, (\lambda) = 500 nm = (5 \times 10^{-7}) m
  • Aperture diameter, (d) = 0.02 mm = (2 \times 10^{-5}) m

• The radius of the first dark ring is given by: $$r_{dark} = \frac{1.22 \times \lambda}{d}$$

• Substituting the values, we get: $$r_{dark} = \frac{1.22. \times 5 \times 10^{-7} \text{ m} }{2 \times 10^{-5} \text{ m}}$$ $$r_{dark} = 3.05 \times 10^{-5}\text{ m}$$

Therefore, the radius of the first dark ring is (3.05 \times 10^{-5}) m.

3. Problem: A grating has 500 lines per mm. Calculate the angle at which the first-order diffraction maximum is observed for light of wavelength 600 nm.

Solution:

• Given:

  • Grating lines, (N) = 500 lines/mm
  • Wavelength, (\lambda) = 600 nm = (6 \times 10^{-7}) m

• The angle (\theta_1) at which the first-order diffraction maximum is observed is given by: $$\sin \theta_1 = \frac{\lambda}{d}$$ where (d) is the grating spacing.

• The grating spacing (d) is given by: $$d = \frac{1}{N}$$

• Substituting the value of (d) into the first equation, we get: $$\sin \theta_1 = \lambda N$$ $$\theta_1 = \sin^{-1}(\lambda N)$$

• Substituting the values, we get: $$\theta_1 = \sin^{-1}(6 \times 10^{-7} \text{ m} \times 500 \text{ lines/mm})$$ $$\theta_1 = 18.43\degree$$

Therefore, the angle at which the first-order diffraction maximum is observed is (18.43\degree).

4. Problem: A beam of light of wavelength 650 nm is incident on a grating with 200 lines per mm. Calculate the number of bright fringes observed on a screen placed 1 m away from the grating.

Solution:

• Given:

  • Wavelength, (\lambda) = 650 nm = (6.5 \times 10^{-7}) m
  • Grating lines, (N) = 200 lines/mm
  • Distance from grating to screen, (D) = 1 m

• The angular width (\theta) of each bright fringe is given by: $$\theta = \frac{\lambda}{d}$$ where (d) is the grating spacing.

• The grating spacing (d) is given by: $$d = \frac{1}{N}$$

• Substituting the value of (d) into the first equation, we get: $$\theta = \lambda N$$

• The total number of bright fringes observed on the screen is given by: $$m = \frac{D}{\tan \theta}$$ where (m) is the number of bright fringes and (D) is the distance from the grating to the screen.

• Substituting the values, we get: $$m = \frac{1 \text{ m}}{\tan (6.5 \times 10^{-7} \text{ m} \times 200 \text{ lines/mm})}$$ $$m = 4.4$$

Therefore, the number of bright fringes observed on the screen is (4).

5. Problem: A parallel beam of light of wavelength 550 nm falls on a double slit of width 0.01 mm and separation 0.02 mm. Calculate the angular separation between the two central maxima in the diffraction pattern.

Solution:

• Given:

  • Wavelength, (\lambda) = 550 nm = (5.5 \times 10^{-7}) m
  • Slit width, (a) = 0.01 mm = (1 \times 10^{-5}) m
  • Slit separation, (d) = 0.02 mm = (2 \times 10^{-5}) m

• The angular separation (\theta) between the two central maxima is given by: $$\theta = \frac{\lambda d}{a}$$

• Substituting the values, we get: $$\theta = \frac{5.5 \times 10^{-7} \text{ m} \times 2 \times 10^{-5} \text{ m}}{1 \times 10^{-5} \text{ m}}$$ $$\theta = 1.1 \times 10^{-1}\text{ radians}$$ $$\theta = 6.34\degree$$

Therefore, the angular separation between the two central maxima is (6.34\degree).

CBSE Board Exams Numerical

1. Problem: A parallel beam of light of wavelength 600 nm is incident on a single slit of width 0.01 mm. Calculate the angle at which the first dark fringe is formed.

Solution:

• Given:

  • Wavelength, (\lambda) = 600 nm = (6 \times 10^{-7}) m
  • Slit width, (a) = 0.01 mm = (1 \times 10^{-5}) m

• The angle (\theta) at which the first dark fringe is formed is given by: $$\sin \theta = \frac{\lambda}{a}$$

• Substituting the values, we get: $$\sin \theta = \frac{6 \times 10^{-7} \text{ m} }{1 \times 10^{-5} \text{ m}}$$ $$\theta = \sin^{-1}(6 \times 10^{-2})$$ $$\theta = 3.43\degree$$

Therefore, the angle at which the first dark fringe is formed is (3.43\degree).

2. Problem: A monochromatic light of wavelength 600 nm is incident on a circular aperture of diameter 0.02 mm. Calculate the diameter of the central bright spot in the diffraction pattern.

Solution:

• Given:

  • Wavelength, (\lambda) = 600 nm = (6 \times 10^{-7}) m
  • Aperture diameter, (d) = 0.02 mm = (2 \times 10^{-5}) m

• The diameter of the central bright spot is given by: $$D = \frac{2.44\lambda f}{d}$$

  where (f) is the focal length of the lens used to observe the diffraction pattern.

• Assuming (f) is very large (lens placed at a large distance), the diameter of the central spot can be approximated by: $$D \approx 2.44 \times \lambda$$

• Substituting the value of (\lambda), we get: $$D \approx 2.44 \times 6 \times 10^{-7} \text{ m}$$ $$D \approx 1.464 \times 10^{-6}\text{ m}$$

**Therefore, the diameter of the central bright spot is approximately (1.46 \times 10