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In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is

Answers (1)

As we learnt in

Young Double Slit Experiment -

- wherein

y=\Delta x\cdot \left ( \frac{D}{d} \right )

y= Distance of a point on screen from central maxima

\Delta x= Path difference at that point

 

 

Ref y be the distance from the central maximum to the point where the bright fringes due to both the wavelength coincides

For \lambda_{1},  y=\frac{m\lambda_{1}D}{d}, 

for  \lambda_{2},   y=\frac{n\lambda_{2}D}{d}

\therefore\ \; m \lambda_{1}=n\lambda_{2}

\Rightarrow\ \; \frac{m}{n}= \frac{\lambda_{2}}{\lambda_{1}}=\frac{620}{650}=\frac{4}{5}

i.e. with respect to central maximum 4th bright fringe of \lambda_{1} coincides with 5th bright fringe of \lambda_{2}

y=\frac{4\times 650\times 10^{-9}\times 1.5}{0.5\times 10^{-3}}m

y=7.8\times10^{-3}m   or     y = 7.8 mm

Posted by

Abhishek Sahu

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