Using Young’s double slit experiment, a monochromatic light of wavelength 5000 $\overset{o}{A}$ produces fringes of fringe width 0.5mm. If another monochromatic light of wavelength 6000 $\overset{o}{A}$ is used and the separation between the slits is doubled, then the new fringe width will be:

Option 1 -

0.5 mm

Option 2 -

1 mm

Option 3 -

0.6 mm

Option 4 -

0.3 mm

2 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$β_{1} = \frac{λ_{1} D_{1}}{d_{1}}$

$β_{2} = \frac{λ_{2} D_{2}}{d_{2}}$

$\frac{β_{1}}{β_{2}} = \frac{λ_{1}}{d_{1}} \times \frac{d_{2}}{λ_{2}}$

$\frac{0.5}{β_{2}} = \frac{5000 \times d_{1}}{d_{1} \times 600}$

$β_{2} = \frac{0.5 \times 6}{10}$

2 = 0.3mm

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Using Young’s double slit experiment, a monochromatic light of wavelength 5000 $\overset{o}{A}$ produces fringes of fringe width 0.5mm. If another monochromatic light of wavelength 6000 $\overset{o}{A}$ is used and the separation between the slits is doubled, then the new fringe width will be:

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Using Young’s double slit experiment, a monochromatic light of wavelength 5000 A o produces fringes of fringe width 0.5mm. If another monochromatic light of wavelength 6000 A o is used and the separation between the slits is doubled, then the new fringe width will be:

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Using Young’s double slit experiment, a monochromatic light of wavelength 5000 $\overset{o}{A}$ produces fringes of fringe width 0.5mm. If another monochromatic light of wavelength 6000 $\overset{o}{A}$ is used and the separation between the slits is doubled, then the new fringe width will be: