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NEET
Exam
NEET
Subject
Physics
Chapter
Wave Optics

NEET · Physics · Wave Optics

In a Young's double slit experiment, the slits are separated by 0.3...

In a Young's double slit experiment, the slits are separated by 0.3 mm0.3 \mathrm{~mm} and the screen is placed 1.5 m1.5 \mathrm{~m} away. If the measured fringe width is 2 mm2 \mathrm{~mm}, the wavelength of the light used is

Options

  1. A4000 nm4000 \mathrm{~nm}
  2. B250 nm250 \mathrm{~nm}
  3. C400 nm400 \mathrm{~nm}
  4. D1000 nm1000 \mathrm{~nm}

Correct answer

C. 400 nm400 \mathrm{~nm}

Step-by-step solution

Fringe width is given by β=λDd\beta = \dfrac{\lambda D}{d}

Rearranging for wavelength λ\lambda:
λ=βdD\lambda = \dfrac{\beta d}{D}

Substitute the given values:
β=2 mm=2×10−3 m\beta = 2 \mathrm{~mm} = 2 \times 10^{-3} \mathrm{~m}
d=0.3 mm=0.3×10−3 md = 0.3 \mathrm{~mm} = 0.3 \times 10^{-3} \mathrm{~m}
D=1.5 mD = 1.5 \mathrm{~m}

λ=2×10−3×0.3×10−31.5\lambda = \dfrac{2 \times 10^{-3} \times 0.3 \times 10^{-3}}{1.5}
λ=0.6×10−61.5=0.4×10−6 m\lambda = \dfrac{0.6 \times 10^{-6}}{1.5} = 0.4 \times 10^{-6} \mathrm{~m}

Converting to nanometers:
λ=400×10−9 m=400 nm\lambda = 400 \times 10^{-9} \mathrm{~m} = 400 \mathrm{~nm}

Answer: 400 nm400 \mathrm{~nm}
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