Two sources of light $0.6 \mathrm{~mm}$ apart and screen is placed at a distance of $1.2 \mathrm{~m}$ from…

Two sources of light $0.6 \mathrm{~mm}$ apart and screen is placed at a distance of $1.2 \mathrm{~m}$ from them. A light of wavelength $6000 Å$ used. Then the phase difference between the two light waves interfering on the screen at a point at a distance $3 \mathrm{~mm}$ from central bright band is
  1. $6 \pi$ radian
  2. $3 \pi$ radian
  3. $4 \pi$ radian
  4. $5 \pi$ radian

Solution

Fringe width, $W=\frac{\lambda . D}{d}=\frac{6000 \times 10^{-10} \times 1.2}{0.6 \times 10^{-3}}=1.2 \mathrm{~mm}$ $\therefore \quad$ Number of fringes (n) $=\frac{3}{1.2}=2.5$ $\therefore \quad$ Phase difference, $\Delta \phi=2 \mathrm{n} \pi=2 \times 2.5 \pi=5 \pi$

Asked in: MHT CET 2023 (10 May Shift 1)

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