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Wave Optics question

2021 · 16 Mar · Shift 1 · Q63
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Wave Optics question

2021 · 16 Mar · Shift 1 · Q63

JEE MainPhysicsWave OpticsNumerical+4 / −1
A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 600

  1. For Young’s double slit experiment, the fringe width is β=λDd\beta = \frac{\lambda D}{d}β=dλD​ where:

    • β=6 mm=6×10−3 m\beta = 6\,\text{mm} = 6 \times 10^{-3}\,\text{m}β=6mm=6×10−3m
    • d=1 mm=1×10−3 md = 1\,\text{mm} = 1 \times 10^{-3}\,\text{m}d=1mm=1×10−3m
    • D=10 mD = 10\,\text{m}D=10m

  2. Rearranging for wavelength: λ=βdD\lambda = \frac{\beta d}{D}λ=Dβd​

  3. Substitute the values: λ=(6×10−3)(1×10−3)10\lambda = \frac{(6 \times 10^{-3})(1 \times 10^{-3})}{10}λ=10(6×10−3)(1×10−3)​

  4. Compute: λ=6×10−610=6×10−7 m\lambda = \frac{6 \times 10^{-6}}{10} = 6 \times 10^{-7}\,\text{m}λ=106×10−6​=6×10−7m

  5. Convert to nanometres: 1 nm=10−9 m1\,\text{nm} = 10^{-9}\,\text{m}1nm=10−9m so λ=6×10−7 m=600 nm\lambda = 6 \times 10^{-7}\,\text{m} = 600\,\text{nm}λ=6×10−7m=600nm

  6. Therefore, the required integer value is: x=600x = 600x=600

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