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Exercises · 10.4

Q.In a Young's double-slit experiment, the slits are separated by 0.28 mm0.28\ \text{mm} and the screen is placed 1.4 m1.4\ \text{m} away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm1.2\ \text{cm}. Determine the wavelength of light used in the experiment.

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In Young’s double-slit interference, the bright fringe positions are given by yn=nλDdy_n = n \frac{\lambda D}{d}. Using the given values for the fourth bright fringe, the wavelength is found to be λ=600 nm\lambda = 600\ \text{nm}.

The key to solving this problem is understanding that in Young’s double-slit experiment, bright fringes (constructive interference) occur at positions where the path difference from the two slits is an integer multiple of the wavelength. The formula yn=nλDdy_n = n \frac{\lambda D}{d} directly relates the fringe position to the wavelength, slit separation, and screen distance. Here, we know the distance to the fourth bright fringe (n=4n=4), so we can solve for λ\lambda directly.

Let’s work through it step by step.

  1. Identify the known quantities.

    Slit separation: d=0.28 mm=0.28×10−3 m=2.8×10−4 md = 0.28\ \text{mm} = 0.28 \times 10^{-3}\ \text{m} = 2.8 \times 10^{-4}\ \text{m}.

    Screen distance: D=1.4 mD = 1.4\ \text{m}.

    Distance to the fourth bright fringe from the central maximum: y4=1.2 cm=1.2×10−2 my_4 = 1.2\ \text{cm} = 1.2 \times 10^{-2}\ \text{m}.

    Fringe order: n=4n = 4.

  2. Recall the formula for bright fringe positions.

    For constructive interference in Young’s double-slit, the nn-th bright fringe (where n=0,1,2,…n = 0, 1, 2, \dots) is located at a distance from the central maximum given by:

yn=nλDdy_n = n \frac{\lambda D}{d}

Here n=0n=0 gives the central bright fringe, n=1n=1 the first bright fringe, and so on. The problem states “the distance between the central bright fringe and the fourth bright fringe” — that is exactly y4y_4.

  1. Substitute the known values into the formula.

1.2×10−2=4×λ×1.42.8×10−41.2 \times 10^{-2} = 4 \times \frac{\lambda \times 1.4}{2.8 \times 10^{-4}}

  1. Solve for λ\lambda. First, simplify the right-hand side:

4×1.42.8×10−4λ=5.62.8×10−4λ=2×104λ\frac{4 \times 1.4}{2.8 \times 10^{-4}} \lambda = \frac{5.6}{2.8 \times 10^{-4}} \lambda = 2 \times 10^{4} \lambda

So the equation becomes:

1.2×10−2=2×104λ1.2 \times 10^{-2} = 2 \times 10^{4} \lambda

Divide both sides by 2×1042 \times 10^{4}:

λ=1.2×10−22×104=0.6×10−6 m=6×10−7 m\lambda = \frac{1.2 \times 10^{-2}}{2 \times 10^{4}} = 0.6 \times 10^{-6}\ \text{m} = 6 \times 10^{-7}\ \text{m} …

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