In a Young's double slit experiment, the fringes are obtained on a screen at distance D from the slits. The separation between the slits is d and the wavelength of light is λ. Arrange the fringe width in descending order for the situations given below:
(A) D = 1.5 m, d = 0.15 mm and λ = 500 nm
(B) D = 1.2 m, d = 0.1 mm and λ = 600 nm
(C) D = 1.2 m, d = 0.04 mm and λ = 500 nm
(D) D = 1.0 m, d = 0.05 mm and λ = 500 nm
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (C), (D), (B), (A)
The fringe width in Young’s double slit experiment is directly proportional to \( \lambda \) and D and inversely proportional to the slit separation d. Convert every length to metres.
\( \displaystyle \beta = \frac{\lambda D}{d} \)
(A) D = 1.5 m, d = 0.15 mm and λ = 500 nm — \( \displaystyle \beta_A = \frac{500\times10^{-9}\,\mathrm{m} \times 1.5\,\mathrm{m}}{0.15\times10^{-3}\,\mathrm{m}} = 5\times10^{-3}\,\mathrm{m} \)
(B) D = 1.2 m, d = 0.1 mm and λ = 600 nm — \( \displaystyle \beta_B = \frac{600\times10^{-9}\,\mathrm{m} \times 1.2\,\mathrm{m}}{0.1\times10^{-3}\,\mathrm{m}} = 7.2\times10^{-3}\,\mathrm{m} \)
(C) D = 1.2 m, d = 0.04 mm and λ = 500 nm — \( \displaystyle \beta_C = \frac{500\times10^{-9}\,\mathrm{m} \times 1.2\,\mathrm{m}}{0.04\times10^{-3}\,\mathrm{m}} = 15\times10^{-3}\,\mathrm{m} \)
(D) D = 1.0 m, d = 0.05 mm and λ = 500 nm — \( \displaystyle \beta_D = \frac{500\times10^{-9}\,\mathrm{m} \times 1.0\,\mathrm{m}}{0.05\times10^{-3}\,\mathrm{m}} = 10\times10^{-3}\,\mathrm{m} \)
So the fringe widths are 15 mm (C), 10 mm (D), 7.2 mm (B) and 5 mm (A), and the descending order is (C), (D), (B), (A).
Shortcut: Only the order is needed, so the unit conversions can be skipped: the same factor applies to all four cases. Compare \( \lambda D/d \) with \( \lambda \) in nm, D in m and d in mm: 5000 (A), 7200 (B), 15000 (C), 10000 (D).