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Exercises · 10.6

Q.A beam of light consisting of two wavelengths, 650 nm650\ \text{nm} and 520 nm520\ \text{nm}, is used to obtain interference fringes in a Young's double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm650\ \text{nm}.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
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Using the standard NCERT data for this problem (D=1.2 mD=1.2\ \text{m}, d=2 mmd=2\ \text{mm}, not repeated in the stored stem): (a) the third bright fringe for 650 nm650\ \text{nm} is at y3≈1.17 mmy_3\approx1.17\ \text{mm} from the centre;

(b) the bright fringes of the two wavelengths first coincide at y≈1.56 mmy\approx1.56\ \text{mm}.

The governing relation

In Young's double-slit experiment, a bright fringe occurs at path difference dsin⁡θ=nλd\sin\theta=n\lambda; for small angles (sin⁡θ≈y/D\sin\theta\approx y/D), the nn-th bright fringe sits at

yn=nλDd.y_n = \frac{n\lambda D}{d}.

(This problem is the standard NCERT exercise, which supplies slit separation d=2 mmd=2\ \text{mm} and screen distance D=1.2 mD=1.2\ \text{m}; these values are used below.)

(a) Third bright fringe for λ1=650 nm\lambda_1=650\ \text{nm}

The central maximum is n=0n=0, so the third bright fringe is n=3n=3:

y3=3λ1Dd=3×(650×10−9)×1.22×10−3=2.34×10−62×10−3≈1.17×10−3 m=1.17 mm.y_3 = \frac{3\lambda_1 D}{d} = \frac{3\times(650\times10^{-9})\times1.2}{2\times10^{-3}} = \frac{2.34\times10^{-6}}{2\times10^{-3}} \approx 1.17\times10^{-3}\ \text{m} = 1.17\ \text{mm}.

(b) Least distance where bright fringes of both wavelengths coincide

A bright fringe of λ1=650 nm\lambda_1=650\ \text{nm} lands on a bright fringe of λ2=520 nm\lambda_2=520\ \text{nm} when their positions match:

n1λ1=n2λ2⟹n1(650)=n2(520)⟹n1n2=520650=45.n_1\lambda_1 = n_2\lambda_2 \quad\Longrightarrow\quad n_1(650) = n_2(520) \quad\Longrightarrow\quad \frac{n_1}{n_2}=\frac{520}{650}=\frac{4}{5}.

The smallest positive integers satisfying this are n1=4n_1=4, n2=5n_2=5 (i.e. the 4th bright fringe of the 650 nm650\ \text{nm} light coincides with the 5th bright fringe of the 520 nm520\ \text{nm} light - check: 4×650=2600=5×5204\times650=2600=5\times520, confirmed). …

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