Interference and Diffraction: Speed-Solving Techniques (4)

easy • 2 min read
Tags Interference And Diffraction

Question

In Young’s double-slit experiment, the slit separation is d=0.5 mmd = 0.5\ \text{mm}, the screen is D=1 mD = 1\ \text{m} away, and light of wavelength λ=500 nm\lambda = 500\ \text{nm} is used. Find the fringe width in millimetres in under 30 seconds.

Solution — Step by Step

β=λDd\beta = \frac{\lambda D}{d}

λ=500×10−9 m\lambda = 500 \times 10^{-9}\ \text{m}, D=1 mD = 1\ \text{m}, d=0.5×10−3 md = 0.5 \times 10^{-3}\ \text{m}.

 β=500×10−9×10.5×10−3=10−3 m=1 mm\beta = \frac{500 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 10^{-3}\ \text{m} = 1\ \text{mm}

Final answer: β=1 mm\beta = 1\ \text{mm}.

Why This Works

YDSE problems boil down to one formula. The fringe width β\beta is constant across the pattern — bright and dark fringes are equally spaced. Once you’ve memorised β=λD/d\beta = \lambda D/d, the question is just unit conversion.

The deeper geometry: path difference =dsin⁡θ≈dθ=dx/D= d\sin\theta \approx d\theta = dx/D. Setting this equal to nλn\lambda gives xn=nλD/dx_n = n\lambda D/d, so consecutive maxima are spaced by λD/d\lambda D/d.

Alternative Method

If only the position of the nn-th bright fringe is asked, use xn=nλD/dx_n = n\lambda D/d. For dark fringes, xn=(n+12)λD/dx_n = (n + \tfrac{1}{2})\lambda D/d.

If the experiment is immersed in water (refractive index μ\mu), wavelength becomes λ/μ\lambda/\mu, so fringe width also shrinks by μ\mu. JEE Main 2023 PYQ used water with μ=4/3\mu = 4/3.

Common Mistake

Mixing units. Writing dd in mm and DD in metres without converting gives an answer off by 1000. Always go full SI for one calculation, then convert the final answer once.

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