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Wave Optics question

2021 · 18 Mar · Shift 1 · Q53
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Wave Optics question

2021 · 18 Mar · Shift 1 · Q53

JEE MainPhysicsWave OpticsMCQ+4 / −1
In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 Ao\mathop A\limits^oAo​ is :-
  1. A
    1178 ×\times× 10 −-− 9 m
  2. B
    1178 ×\times× 10 −-− 6 m
  3. C
    1178 ×\times× 10 −-− 12 m
  4. D
    5890 ×\times× 10 −-− 7 m
View written solutionFree

Correct answer: B

  1. Given data

    d=0.5 mm=0.5×10−3 md = 0.5\ \text{mm} = 0.5 \times 10^{-3}\ \text{m}d=0.5 mm=0.5×10−3 m D=0.5 mD = 0.5\ \text{m}D=0.5 m λ=5890 A˚=5890×10−10 m\lambda = 5890\ \mathring{A} = 5890 \times 10^{-10}\ \text{m}λ=5890 A˚=5890×10−10 m

  2. Fringe width in Young's double slit experiment

    The fringe width is β=λDd\beta = \frac{\lambda D}{d}β=dλD​

    Substituting the values: β=(5890×10−10)(0.5)0.5×10−3\beta = \frac{(5890 \times 10^{-10})(0.5)}{0.5 \times 10^{-3}}β=0.5×10−3(5890×10−10)(0.5)​

    Since 0.50.50.5 cancels out, β=5890×10−10+3\beta = 5890 \times 10^{-10+3}β=5890×10−10+3 β=5890×10−7 m\beta = 5890 \times 10^{-7}\ \text{m}β=5890×10−7 m β=5.89×10−4 m\beta = 5.89 \times 10^{-4}\ \text{m}β=5.89×10−4 m

  3. Position of bright fringes

    Bright fringes occur at yn=nβy_n = n\betayn​=nβ where n=0,1,2,3,…n=0,1,2,3,\dotsn=0,1,2,3,…

    • First bright fringe from the central bright fringe: y1=βy_1 = \betay1​=β
    • Third bright fringe from the central bright fringe: y3=3βy_3 = 3\betay3​=3β

    Therefore, the distance between the first and third bright fringes is y3−y1=3β−β=2βy_3 - y_1 = 3\beta - \beta = 2\betay3​−y1​=3β−β=2β

  4. Compute the required distance

    2β=2(5890×10−7)2\beta = 2(5890 \times 10^{-7})2β=2(5890×10−7) =11780×10−7= 11780 \times 10^{-7}=11780×10−7 =1178×10−6 m= 1178 \times 10^{-6}\ \text{m}=1178×10−6 m

  5. Matching with options

    1178×10−6 m1178 \times 10^{-6}\ \text{m}1178×10−6 m corresponds to Option B.

Final Answer: Option B

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