Q146 / 200 · Manipal (MET) 2012 · Physics

In a Young's double slit experiment, the separation between the two slits is 0.9 mm and the fringes are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of the monochromatic source of light used is

  • A
    450 nm
  • B
    400 nm
  • C
    500 nm
  • Correct answer
    D
    600 nm

Solution

The distance between two consecutive dark or bright fringes is recognised as $\beta$ (fringe width) and that between central fringe and first dark fringe on either side is $\frac{\beta}{2}$. Given, spacing between second dark fringe and central fringe $=\beta+\frac{\beta}{2}$ or $\frac{3 \beta}{2}=1 \mathrm{~mm}$ (Given) or $\quad \beta=\frac{2}{3} \times 1 \mathrm{~mm}$ or $\frac{\lambda D}{d}=\frac{2}{3} \mathrm{~mm}$ $\begin{array}{ll} \therefore & \lambda=\frac{2}{3} \times 10^{-3} \times \frac{0.9 \times 10^{-3}}{1} \\ \Rightarrow & \lambda=0.6 \times 10^{-6} \mathrm{~m} \end{array}$ $\therefore \quad$ Wavelength of the monochromatic source of light $\lambda=600 \times 10^{-9} \mathrm{~m}=600 \mathrm{~nm}$

Hence, the correct answer is option: 'D'.

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