In Young’s double slit experiment with a monochromatic light of wavelength 4000A0, the fringe width is found to be 0.4 mm. When the slits are now illuminated with a light of wavelength 5000A0 the fringe width will be

λ1=4000A0,β1=0.4mm,λ2=5000A0β2=?β2β1=λ2λ1⇒β2=50004000×0.4=0.5mm

In Young’s double slit experiment with a monochromatic light of wavelength 4000A0, the fringe width is found to be 0.4 mm. When the slits are now illuminated with a light of wavelength 5000A0 the fringe width will be

λ1=4000A0,β1=0.4mm,λ2=5000A0β2=?β2β1=λ2λ1⇒β2=50004000×0.4=0.5mm

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In Young’s double slit experiment with a monochromatic light of wavelength 4000A⁰, the fringe width is found to be 0.4 mm. When the slits are now illuminated with a light of wavelength 5000A⁰ the fringe width will be

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0.5mm

0.32mm

0.6mm

0.8mm

answer is B.

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Detailed Solution

$\begin{array}{l} λ_{1} = 4000 A^{0}, β_{1} = 0 .4 mm, λ_{2} = 5000 A^{0} \\ β_{2} = ? \frac{β_{2}}{β_{1}} = \frac{λ_{2}}{λ_{1}} \Rightarrow β_{2} = \frac{5000}{4000} \times 0 .4 = 0 .5 mm \end{array}$

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