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    a double-slit experiment is performed with light of wavelength 600 nm . the bright interference fringes are spaced 1.8 mm apart on the viewing screen.

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    Find step-by-step Physics solutions and your answer to the following textbook question: A double-slit experiment is performed with light of wavelength 600 nm. The bright interference fringes are spaced 1.8 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 400 nm?.

    Question

    A double-slit experiment is performed with light of wavelength 600 nm. The bright interference fringes are spaced 1.8 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 400 nm?

    Explanation

    Verified \textbf{Given:} Given:

    \color{#4257b2}\bullet \bullet

    ∙∙ \lambda_1=600 λ 1 ​ =600 nm

    =\color{#c34632}600\times10^{-6}

    =600×10 −6 m

    \color{#4257b2}\bullet \bullet

    ∙∙ \Delta y_1=1.8 Δy 1 ​ =1.8 mm

    =\color{#c34632}1.8\times10^{-3}

    =1.8×10 −3 m

    \color{#4257b2}\bullet \bullet

    ∙∙ \lambda_2=400 λ 2 ​ =400 nm

    =\color{#c34632}400\times10^{-6}

    =400×10 −6 m

    \color{#4257b2}\bullet \bullet

    ∙∙ d_1=d_2 d 1 ​ =d 2 ​

    \color{#4257b2}\bullet \bullet

    ∙∙ L_1=L_2 L 1 ​ =L 2 ​

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    A double

    Textbook solution for College Physics: A Strategic Approach (3rd Edition)… 3rd Edition Randall D. Knight (Professor Emeritus) Chapter 17 Problem 10P. We have step-by-step solutions for your textbooks written by Bartleby experts!

    Textbook Question

    Chapter 17, Problem 10P

    A double-slit experiment is performed with light of wavelength 600 nm. The bright interference fringes are spaced 1.8 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 400 nm?

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    Chapter 17, Problem 9P

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    Chapter 17, Problem 11P

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    A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 pm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will :

    Click here👆to get an answer to your question ✍️ A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 pm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will :

    Question

    A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 pm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will :

    A

    Remain unstated

    B

    Shift downward by neary two fringes

    C

    Shift upward by nearly two fringes

    D

    Shift downward by its fringes

    Medium Open in App Solution Verified by Toppr

    Correct option is C)

    In Young's Double Slit Experiment (YDSE) be the position of central bright fringe on the screen in absence of any film, because the S

    1 ​ O=S 2 ​ O.

    When we introduce a film of thickness t and refractive index p, an additional path difference equal to (μt−t)=(μ−1)t is introduced. The optical path of upper bean becomes longer. For path difference on screen to be zero, path from lower slit S

    2 ​

    should also be more. Thus, central bright fringe wire be located some what at O' as S

    2 ​ O ′ >S 2 ​ O.

    ∴ The fringe pattern shifts upwards.

    Now as a change in path difference of λ corresponds to a change in position on the screen by β=

    d D ​ λ

    Change in optical path difference Δy=(μ−1)t×

    d D ​ Δy=(1.5−1)(2×10 −6 ) d D ​ Δy= 2 1 ​ ×2×10 −6 d D ​ = d D ​ ×10 −6 m Fringe width = d D ​ λ= d D ​ ×100×10 −9 m=β Clearly, β Δy ​ = d D ​ ×500×10 −9 d D ​ ×10 −6 ​ =2 = 500×10 −9 10 −6 ​ = 500 10 −6 ×10 9 ​ = 500 10 3 ​ ∴Δy=2β Video Explanation

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