Question 345: Waves Optics

Question

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 $$6000\,\mathop A\limits^o$$ 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

Accepted Answer

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

Answer

$$6 \pi$$ radian

Answer

$$3 \pi$$ radian

Answer

$$4 \pi$$ radian

Answer

$$5 \pi$$ radian

Waves Optics

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