Q85 / 100 · BITSAT 2024 (Memory Based Paper 3) · Physics

In YDSE, monochromatic light falls on a screen 1.80 m from two slits separated by 2.08 mm . The first and second order bright fringes are separated by 0.553 mm . The wavelength of light used is

  • A
    520 nm
  • Correct answer
    B
    639 nm
  • C
    715 nm
  • D
    None of these

Solution

Position of first bright fringe, $y_1=\frac{\lambda D}{d}$ Position of second bright fringe, $\begin{aligned} y_2 & =\frac{2 \lambda D}{d} \\ y_2-y_1 & =\frac{2 \lambda D}{d}-\frac{\lambda D}{d} \\ y_2-y_1 & =\frac{\lambda D}{d} \\ or \lambda & =\frac{\left(y_2-y_1\right) d}{D} \end{aligned}$ Substituting given values, we have $\begin{aligned} \lambda & =\frac{0.553 \times 10^{-3} \times 2.08 \times 10^{-3}}{1.8} \\ & =639 \times 10^{-9} \mathrm{~m} \\ or \lambda & =639 \mathrm{~nm} \end{aligned}$

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

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