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: $$\theta =\frac{2\times 6\times {10}^{-7}\text{ m}}{5\times {10}^{-4}\text{ m}}$$ $$\theta =2.4\times {10}^{-3}\text{ radians}$$ $$\theta =1.37\text{degree}$$

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: $${\theta }_1={\sin }^{-1}(6\times {10}^{-7}\text{ m}\times 500\text{ lines/mm})$$ $${\theta }_1=18.43\text{degree}$$

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: $$\theta =\frac{5.5\times {10}^{-7}\text{ m}\times 2\times {10}^{-5}\text{ m}}{1\times {10}^{-5}\text{ m}}$$ $$\theta =1.1\times {10}^{-1}\text{ radians}$$ $$\theta =6.34\text{degree}$$

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: $$\sin \theta =\frac{6\times {10}^{-7}\text{ m}}{1\times {10}^{-5}\text{ m}}$$ $$\theta ={\sin }^{-1}(6\times {10}^{-2})$$ $$\theta =3.43\text{degree}$$

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