In YDSE slit separation 0.5mm wavelength 500nm — find fringe width at 1m

easy CBSE JEE-MAIN NEET • 3 min read
Tags Interference And Diffraction

Question

In Young’s Double Slit Experiment (YDSE), the slit separation is d=0.5d = 0.5 mm and the wavelength of light is λ=500\lambda = 500 nm. The screen is placed at a distance D=1D = 1 m from the slits. Find the fringe width.

Solution — Step by Step

In YDSE, bright fringes (constructive interference) occur where the path difference is an integer multiple of λ\lambda. The position of the nn-th bright fringe from the central maximum is:

yn=nλDdy_n = \frac{n\lambda D}{d}

The fringe width β\beta is the distance between consecutive bright (or dark) fringes:

β=yn+1−yn=λDd\beta = y_{n+1} - y_n = \frac{\lambda D}{d}
  • d=0.5 mm=0.5×10−3 m=5×10−4d = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m} = 5 \times 10^{-4} m
  • λ=500 nm=500×10−9 m=5×10−7\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} m
  • D=1D = 1 m

Unit consistency is the most important step in this type of problem.

 β=λDd=5×10−7×15×10−4\beta = \frac{\lambda D}{d} = \frac{5 \times 10^{-7} \times 1}{5 \times 10^{-4}} β=5×10−75×10−4=10−7−(−4)=10−3 m\beta = \frac{5 \times 10^{-7}}{5 \times 10^{-4}} = 10^{-7-(-4)} = 10^{-3} \text{ m} β=1 mm\boxed{\beta = 1 \text{ mm}}

The fringes are spaced 1 mm apart on the screen.

Why This Works

Young’s double slit creates two coherent sources (the two slits) that emit waves in phase. At any point on the screen, the waves from the two slits travel different path lengths. When the path difference is 0,λ,2λ,…0, \lambda, 2\lambda, \ldots (integer multiples), the waves arrive in phase → constructive interference → bright fringe. When path difference is λ/2,3λ/2,…\lambda/2, 3\lambda/2, \ldots, they arrive out of phase → destructive interference → dark fringe.

The fringe formula β=λD/d\beta = \lambda D/d shows three key relationships:

  • Larger λ\lambda → wider fringes (red light gives wider fringes than violet)
  • Larger DD → wider fringes (moving screen farther spreads the pattern)
  • Larger dd → narrower fringes (fringes compress as slits move apart)

Alternative Method

A quick cross-check: β=λD/d\beta = \lambda D / d. If λ\lambda and DD are both increased by the same factor, β\beta increases by that factor. If dd doubles, β\beta halves. Use these relationships to mentally verify your answer before calculating — if β\beta comes out in kilometres, something’s wrong.

Common Mistake

The most common mistake is mixing units — using λ\lambda in nm and dd in mm and not converting both to the same unit. For example, β=500×10−9/0.5×10−3=500/0.5×10−9+3=1000×10−6=10−3\beta = 500 \times 10^{-9} / 0.5 \times 10^{-3} = 500/0.5 \times 10^{-9+3} = 1000 \times 10^{-6} = 10^{-3} m = 1 mm. If you forget to convert and calculate 500/0.5=1000500/0.5 = 1000 and stop there, you’ve lost the powers of 10. Always write the unit at every step.

JEE Main frequently modifies this basic problem: “What happens to fringe width if the whole apparatus is immersed in water (n=1.33n = 1.33)?” — Answer: λ\lambda in water = λair/n\lambda_{air}/n, so βwater=βair/n=1/1.33≈0.75\beta_{water} = \beta_{air}/n = 1/1.33 \approx 0.75 mm. Fringe width decreases by factor nn.

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