Interference and Diffraction: Application Problems (1)

easy • 3 min read
Tags Interference And Diffraction

Question

In a Young’s double-slit experiment, slits separated by d=0.5 mmd = 0.5 \text{ mm} are illuminated with monochromatic light of wavelength λ=600 nm\lambda = 600 \text{ nm}. The screen is D=1.5 mD = 1.5 \text{ m} away. Find (a) the fringe width, (b) the position of the third bright fringe, and (c) what happens to the fringe width if the entire setup is immersed in water of refractive index 4/34/3.

Solution — Step by Step

For YDSE the fringe width (distance between two consecutive bright or dark fringes) is:

β=λDd\beta = \frac{\lambda D}{d}

Plug in: λ=600×10−9 m\lambda = 600 \times 10^{-9} \text{ m}, D=1.5 mD = 1.5 \text{ m}, d=0.5×10−3 md = 0.5 \times 10^{-3} \text{ m}.

β=(600×10−9)(1.5)0.5×10−3=1.8×10−3 m=1.8 mm\beta = \frac{(600 \times 10^{-9})(1.5)}{0.5 \times 10^{-3}} = 1.8 \times 10^{-3} \text{ m} = 1.8 \text{ mm}

Bright fringes occur at yn=nβy_n = n\beta from the central maximum:

y3=3×1.8=5.4 mmy_3 = 3 \times 1.8 = 5.4 \text{ mm}

Wavelength in a medium decreases by the refractive index: λ′=λ/nw\lambda' = \lambda/n_w. So the new fringe width:

β′=λ′Dd=βnw=1.84/3=1.35 mm\beta' = \frac{\lambda' D}{d} = \frac{\beta}{n_w} = \frac{1.8}{4/3} = 1.35 \text{ mm}

The pattern compresses — fringes get closer together by a factor of 4/34/3.

Why This Works

YDSE depends only on three quantities: source wavelength λ\lambda, slit separation dd, and screen distance DD. The geometry of the setup (small angles, D≫dD \gg d) makes path difference ≈dsin⁡θ≈dy/D\approx d\sin\theta \approx d y/D, which gives the fringe spacing.

When we immerse in water, the frequency of light stays the same (it’s set by the source) but the speed and wavelength both shrink by the refractive index. Slit separation and screen distance are physical quantities — they don’t change. So β\beta shrinks proportionally to λ\lambda.

Alternative Method

Use angular fringe width: θβ=λ/d\theta_\beta = \lambda/d. In air, θβ=600×10−9/(0.5×10−3)=1.2×10−3\theta_\beta = 600 \times 10^{-9}/(0.5 \times 10^{-3}) = 1.2 \times 10^{-3} rad. Linear fringe width: β=θβD=1.8 mm\beta = \theta_\beta D = 1.8 \text{ mm}. Same answer, sometimes faster when only angular separation is asked.

For a quick fringe width estimate in mm: β (mm)=λ(μm)×D(m)/d(mm)\beta \text{ (mm)} = \lambda(\mu\text{m}) \times D(\text{m})/d(\text{mm}). With λ=0.6μm\lambda = 0.6 \mu\text{m}, D=1.5 mD = 1.5 \text{ m}, d=0.5 mmd = 0.5 \text{ mm}: β=0.6×1.5/0.5=1.8\beta = 0.6 \times 1.5 / 0.5 = 1.8. Matches our calculation in seconds.

Common Mistake

The position of the third dark fringe (minimum) is (n+12)β=2.5β(n + \tfrac{1}{2})\beta = 2.5\beta from the centre — not 3β3\beta. Students who don’t read the question carefully report the wrong position. Bright fringes: y=nβy = n\beta. Dark fringes: y=(n+12)βy = (n + \tfrac{1}{2})\beta.

The other slip: students apply the refractive index to DD or dd. Neither changes when immersed. Only λ\lambda changes — that’s the only optical quantity in the formula that depends on the medium.

Final answer: β=1.8 mm\beta = 1.8 \text{ mm}, y3=5.4 mmy_3 = 5.4 \text{ mm}, β′=1.35 mm\beta' = 1.35 \text{ mm} in water.

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