Verified Solution • 2025–26 Session moderate

QNEET-2024-Ph-141: NEET 2024 Physics - Class 12 NEET-2024-Ph-141 Solution: Two slits in Young's double slit experiment are 1.5 mm apart…

Two slits in Young's double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is 600×10−9 m600 \times 10^{-9}\ \text{m} then the fringe separation is 4×10−5 m4 \times 10^{-5}\ \text{m} 9×10−8 m9 \times 10^{-8}\ \text{m} 4×10−7 m4 \times 10^{-7}\ \text{m} 4×10−4 m4 \times 10^{-4}\ \text{m}

Multiple Choice Options

  • A 4×10−5 m4 \times 10^{-5} \mathrm{~m} 4 _ 1 0 _ 5   m
  • B 9×10−8 m9 \times 10^{-8} \mathrm{~m} 9 _ 1 0 _ 8   m
  • C 4×10−7 m4 \times 10^{-7} \mathrm{~m} 4 _ 1 0 _ 7   m
  • D 4×10−4 m4 \times 10^{-4} \mathrm{~m} 4 _ 1 0 _ 4   m

Step-by-Step Verified Solution

To find the fringe separation (also known as fringe width) in Young's double slit experiment, we use the formula: _ = _ D d where: _ is the fringe width (fringe separation). _ is the wavelength of the light used. D is the distance from the slits to the screen. d is the distance between the two slits. Given values are: d = 1.5 mm = 1.5 _ 10 _ 3 m D = 1 m _ = 600 _ 10 _ 9 m Substituting these values into the formula: _ = 600 _ 10 _ 9 m _ 1 m 1.5 _ 10 _ 3 m Calculating the value: _ = 600 _ 10 _ 9 1.5 _ 10 _ 3 _ = 4 _ 10 _ 4 m Thus, the fringe separation is 4 _ 10 _ 4 m . The correct answer is: Option D 4 _ 10 _ 4   m

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