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Wave Optics question

2023 · 6 Apr · Shift 2 · Q73
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Wave Optics question

2023 · 6 Apr · Shift 2 · Q73

JEE MainPhysicsWave OpticsNumerical+4 / −1
A beam of light consisting of two wavelengths 7000 Ao7000~\mathop A\limits^o7000 Ao​ and 5500 Ao5500~\mathop A\limits^o5500 Ao​ is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm2.5 \mathrm{~mm}2.5 mm and the distance between the plane of slits and the screen is 150 cm150 \mathrm{~cm}150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is n×10−5 mn \times 10^{-5} \mathrm{~m}n×10−5 m. The value of nnn is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 462

  1. Fringe width for each wavelength

In Young's double slit experiment, β=λDd\beta = \frac{\lambda D}{d}β=dλD​ where:

  • d=2.5 mm=2.5×10−3 md = 2.5\,\text{mm} = 2.5 \times 10^{-3}\,\text{m}d=2.5mm=2.5×10−3m
  • D=150 cm=1.5 mD = 150\,\text{cm} = 1.5\,\text{m}D=150cm=1.5m

Given wavelengths: λ1=7000 A˚=7×10−7 m\lambda_1 = 7000\,\text{\AA} = 7 \times 10^{-7}\,\text{m}λ1​=7000A˚=7×10−7m λ2=5500 A˚=5.5×10−7 m\lambda_2 = 5500\,\text{\AA} = 5.5 \times 10^{-7}\,\text{m}λ2​=5500A˚=5.5×10−7m

So, β1=7×10−7×1.52.5×10−3=4.2×10−4 m\beta_1 = \frac{7 \times 10^{-7} \times 1.5}{2.5 \times 10^{-3}} = 4.2 \times 10^{-4}\,\text{m}β1​=2.5×10−37×10−7×1.5​=4.2×10−4m β2=5.5×10−7×1.52.5×10−3=3.3×10−4 m\beta_2 = \frac{5.5 \times 10^{-7} \times 1.5}{2.5 \times 10^{-3}} = 3.3 \times 10^{-4}\,\text{m}β2​=2.5×10−35.5×10−7×1.5​=3.3×10−4m

  1. Condition for coincidence of bright fringes

Bright fringes coincide when m1β1=m2β2m_1 \beta_1 = m_2 \beta_2m1​β1​=m2​β2​ for integers m1,m2m_1, m_2m1​,m2​.

That is, m1λ1=m2λ2m_1 \lambda_1 = m_2 \lambda_2m1​λ1​=m2​λ2​ m1×7000=m2×5500m_1 \times 7000 = m_2 \times 5500m1​×7000=m2​×5500 m1m2=55007000=1114\frac{m_1}{m_2} = \frac{5500}{7000} = \frac{11}{14}m2​m1​​=70005500​=1411​

Hence the smallest integers are: m1=11,m2=14m_1 = 11, \quad m_2 = 14m1​=11,m2​=14

  1. Least distance from central fringe

This distance is y=m1β1=11×4.2×10−4y = m_1 \beta_1 = 11 \times 4.2 \times 10^{-4}y=m1​β1​=11×4.2×10−4 y=4.62×10−3 my = 4.62 \times 10^{-3}\,\text{m}y=4.62×10−3m

Now given: y=n×10−5 my = n \times 10^{-5}\,\text{m}y=n×10−5m So, n×10−5=4.62×10−3n \times 10^{-5} = 4.62 \times 10^{-3}n×10−5=4.62×10−3 n=4.62×10−310−5=462n = \frac{4.62 \times 10^{-3}}{10^{-5}} = 462n=10−54.62×10−3​=462

  1. Final answer

462\boxed{462}462​

The derived answer matches the stored correct answer.

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