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Wave Optics question

2020 · 7 Jan · Shift 2 · Q47
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Wave Optics question

2020 · 7 Jan · Shift 2 · Q47

JEE MainPhysicsWave OpticsMCQ+4 / −1
In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :
  1. A
    4.9 mm
  2. B
    5.9 mm
  3. C
    6.9 mm
  4. D
    3.9 mm
View written solutionFree

Correct answer: B

  1. Use the fringe width formula for Young's double slit experiment:

β=λDd\beta = \frac{\lambda D}{d}β=dλD​

where:

  • β\betaβ = separation between successive bright fringes
  • λ=589 nm=589×10−9 m\lambda = 589\,\text{nm} = 589 \times 10^{-9}\,\text{m}λ=589nm=589×10−9m
  • D=1.5 mD = 1.5\,\text{m}D=1.5m
  • d=0.15 mm=0.15×10−3 md = 0.15\,\text{mm} = 0.15 \times 10^{-3}\,\text{m}d=0.15mm=0.15×10−3m
  1. Substitute the values:

β=(589×10−9)(1.5)0.15×10−3\beta = \frac{(589 \times 10^{-9})(1.5)}{0.15 \times 10^{-3}}β=0.15×10−3(589×10−9)(1.5)​

  1. Simplify:

First,

589×1.5=883.5589 \times 1.5 = 883.5589×1.5=883.5

So,

β=883.5×10−90.15×10−3\beta = \frac{883.5 \times 10^{-9}}{0.15 \times 10^{-3}}β=0.15×10−3883.5×10−9​

β=883.50.15×10−6\beta = \frac{883.5}{0.15} \times 10^{-6}β=0.15883.5​×10−6

β=5890×10−6 m\beta = 5890 \times 10^{-6}\,\text{m}β=5890×10−6m

β=5.89×10−3 m\beta = 5.89 \times 10^{-3}\,\text{m}β=5.89×10−3m

  1. Convert to mm:

5.89×10−3 m=5.89 mm5.89 \times 10^{-3}\,\text{m} = 5.89\,\text{mm}5.89×10−3m=5.89mm

  1. Match with the options:

β≈5.9 mm\beta \approx 5.9\,\text{mm}β≈5.9mm

So the correct option is B.

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