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In Young's double slit experiment with a mono-chromatic light of wavelength 4000A^o, the fringe width is found to be 0.4 mm. When the slits are now illuminated with a light of wavelength 5000A^o the fringe width will be

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0.6 mm

0.8 mm

0.5 mm

0.32 mm

answer is B.

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Detailed Solution

{\lambda _1} = 4000{A^0},\,\,{\beta _1} = 0.4mm,\,{\lambda _2} = 5000{A^0}\

{\beta _2} = ?\,\,\frac{{{\beta _2}}}{{{\beta _1}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} \Rightarrow {\beta _2} = \frac{{5000}}{{4000}} \times 0.4 = 0.5\,mm\

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