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Question

# A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0⋅25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.

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Solution

## Given: Wavelengths of the source of light, ${\lambda }_{1}=480×{10}^{-9}\mathrm{m}\mathrm{and}{\lambda }_{2}=600×{10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}$ Separation between the slits, $d=0.25\mathrm{mm}=0.25×{10}^{-3}\mathrm{m}$ Distance between screen and slit, $D=150\mathrm{cm}=1.5\mathrm{m}$ We know that the position of the first maximum is given by $y=\frac{\lambda D}{d}$ So, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = y2 − y1 ${y}_{2}-{y}_{1}=\frac{D\left({y}_{2}-{y}_{1}\right)}{d}$ $⇒{y}_{2}-{y}_{1}=\frac{1.5}{0.25×{10}^{-3}}\left(600×{10}^{-9}-480×{10}^{-9}\right)\phantom{\rule{0ex}{0ex}}{y}_{2}-{y}_{1}=72×{10}^{-5}\mathrm{m}=0.72\mathrm{mm}\phantom{\rule{0ex}{0ex}}$

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