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Wave Optics question

2019 · 9 Jan · Shift 2 · Q60
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Wave Optics question

2019 · 9 Jan · Shift 2 · Q60

JEE MainPhysicsWave OpticsMCQ+4 / −1
In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ\lambdaλ= 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range −-− 30o ≤θ≤\le \theta \le≤θ≤ 30o is :
  1. A
    640
  2. B
    320
  3. C
    321
  4. D
    641
View written solutionFree

Correct answer: D

  1. Condition for bright fringes in YDSE

    Bright fringes occur when the path difference satisfies dsin⁡θ=mλ,d\sin\theta = m\lambda,dsinθ=mλ, where m=0,±1,±2,…m=0, \pm 1, \pm 2, \dotsm=0,±1,±2,…

  2. Given data

    d=0.320 mm=3.20×10−4 md = 0.320\text{ mm} = 3.20\times 10^{-4}\text{ m}d=0.320 mm=3.20×10−4 m λ=500 nm=5.00×10−7 m\lambda = 500\text{ nm} = 5.00\times 10^{-7}\text{ m}λ=500 nm=5.00×10−7 m

  3. Angular range

    We need all bright fringes for −30∘≤θ≤30∘.-30^\circ \le \theta \le 30^\circ.−30∘≤θ≤30∘.

    Since sin⁡30∘=12,\sin 30^\circ = \frac{1}{2},sin30∘=21​, the condition becomes ∣sin⁡θ∣≤12.|\sin\theta| \le \frac{1}{2}.∣sinθ∣≤21​.

    So allowed mmm must satisfy ∣mλd∣≤12.\left|\frac{m\lambda}{d}\right| \le \frac{1}{2}.​dmλ​​≤21​.

  4. Find maximum order

    ∣m∣≤d2λ|m| \le \frac{d}{2\lambda}∣m∣≤2λd​

    Substitute values: d2λ=3.20×10−42×5.00×10−7\frac{d}{2\lambda} = \frac{3.20\times 10^{-4}}{2\times 5.00\times 10^{-7}}2λd​=2×5.00×10−73.20×10−4​

    =3.20×10−41.00×10−6=320= \frac{3.20\times 10^{-4}}{1.00\times 10^{-6}} = 320=1.00×10−63.20×10−4​=320

    Thus, m=−320,−319,…,−1,0,1,…,319,320.m = -320, -319, \dots, -1, 0, 1, \dots, 319, 320.m=−320,−319,…,−1,0,1,…,319,320.

  5. Count total bright fringes

    Number of integers from −320-320−320 to 320320320 inclusive is 320+320+1=641.320 + 320 + 1 = 641.320+320+1=641.

  6. Check options

    • A: 640 ❌
    • B: 320 ❌
    • C: 321 ❌
    • D: 641 ✅

Therefore, the correct answer is D.

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