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anonymous

  • one year ago

Monochromatic light of 605 nm falls on a single slit of width 0.095 mm. The slit is located 85 cm from a screen. How far from the center is the first dark band? A. 0.5 mm B. 5.4 mm C. 540 mm D. 5,400 mm

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  1. rvc
    • one year ago
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    |dw:1434865167255:dw|

  2. Michele_Laino
    • one year ago
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    here the Young's experiment is involved

  3. rvc
    • one year ago
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    i hope i m right with equations

  4. anonymous
    • one year ago
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    ok! what does that mean we do?

  5. rvc
    • one year ago
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    find x

  6. anonymous
    • one year ago
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    what do we plug in? @Michele_Laino @rvc

  7. Michele_Laino
    • one year ago
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    please continue @rvc

  8. rvc
    • one year ago
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    okay

  9. anonymous
    • one year ago
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    what do i plug into the equation @rvc ? :)

  10. rvc
    • one year ago
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    hey the formula

  11. anonymous
    • one year ago
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    yes, but i am not sure what values go where:/

  12. Michele_Laino
    • one year ago
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    according to @rvc we have: \[\large x = \frac{{\lambda d}}{D} = \frac{{605 \times {{10}^{ - 9}} \times 0.095}}{{85}} = ...meters\]

  13. rvc
    • one year ago
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    85 cm

  14. anonymous
    • one year ago
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    ok! so we get 6.76 E-10 ?

  15. Michele_Laino
    • one year ago
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    oops.. \[\large x = \frac{{\lambda d}}{D} = \frac{{605 \times {{10}^{ - 9}} \times 0.095 \times {{10}^{ - 3}}}}{{85 \times {{10}^{ - 2}}}} = ...meters\]

  16. rvc
    • one year ago
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    it should be 0.85 ?

  17. Michele_Laino
    • one year ago
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    yes you are right! @rvc

  18. rvc
    • one year ago
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    thanks

  19. anonymous
    • one year ago
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    ok we get 6.76 E -11 ?

  20. anonymous
    • one year ago
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    i am confused then, what is our solution? :/

  21. Michele_Laino
    • one year ago
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    ok! so how many mm?

  22. anonymous
    • one year ago
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    5,400 mm?

  23. Michele_Laino
    • one year ago
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    you are right! it is not an option of yours @iheartfood

  24. anonymous
    • one year ago
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    ok! so which would i choose? :/

  25. rvc
    • one year ago
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    the formula

  26. Michele_Laino
    • one year ago
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    then try this formula: \[\Large \lambda = \frac{{dx}}{D} \to \quad x = \frac{{\lambda D}}{d}\]

  27. rvc
    • one year ago
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    @Michele_Laino the d

  28. anonymous
    • one year ago
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    please, if you can show me how to use it? i am not sure what to plug in

  29. rvc
    • one year ago
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    yep the formula is wrong :( sorry

  30. Michele_Laino
    • one year ago
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    we have to change the formula @rvc @iheartfood

  31. anonymous
    • one year ago
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    ok! how do we do that?

  32. anonymous
    • one year ago
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    @Michele_Laino ? :)

  33. Michele_Laino
    • one year ago
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    here is the right step: \[\Large x = \frac{{\lambda D}}{d} = \frac{{605 \times {{10}^{ - 9}} \times 85 \times {{10}^{ - 2}}}}{{0.095 \times {{10}^{ - 3}}}} = ...meters\]

  34. rvc
    • one year ago
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    \(\large x = \frac{{\lambda D}}{d} = \frac{{605 \times {{10}^{ - 9}} \times 85 \times {{10}^{ - 2}}}}{{0.095 \times {{10}^{ - 3}}}} = ...meters\)

  35. anonymous
    • one year ago
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    ok! and we get 0.005!

  36. Michele_Laino
    • one year ago
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    I got this: 541315.7*10^(-8) meters

  37. anonymous
    • one year ago
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    ohh ok oops! so +0.5 is the solution?

  38. rvc
    • one year ago
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    541.31 X 10^{-5}

  39. anonymous
    • one year ago
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    i still get 0.005...

  40. Michele_Laino
    • one year ago
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    it is: 54135 * 10^(-5) mm

  41. anonymous
    • one year ago
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    0.54! okay! so our solution is +0.5!

  42. rvc
    • one year ago
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    my ans was in meters

  43. anonymous
    • one year ago
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    yay! thanks both!! onto the next!

  44. Michele_Laino
    • one year ago
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    :)

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