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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Physics
Chapter
Wave Optics

BITSAT 2024 · Physics · Wave Optics

In YDSE, monochromatic light falls on a screen 1.80 m from two...

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

Options

  1. A520 nm
  2. B639 nm
  3. C715 nm
  4. DNone of these

Correct answer

B. 639 nm

Step-by-step solution

Position of first bright fringe, y1=λDdy_1=\frac{\lambda D}{d} Position of second bright fringe, y2=2λDdy2−y1=2λDd−λDdy2−y1=λDdorλ=(y2−y1)dD\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 λ=0.553×10−3×2.08×10−31.8=639×10−9morλ=639nm\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}
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