anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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rvc
  • rvc
|dw:1434865167255:dw|
Michele_Laino
  • Michele_Laino
here the Young's experiment is involved
rvc
  • rvc
i hope i m right with equations

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anonymous
  • anonymous
ok! what does that mean we do?
rvc
  • rvc
find x
anonymous
  • anonymous
what do we plug in? @Michele_Laino @rvc
Michele_Laino
  • Michele_Laino
please continue @rvc
rvc
  • rvc
okay
anonymous
  • anonymous
what do i plug into the equation @rvc ? :)
rvc
  • rvc
hey the formula
anonymous
  • anonymous
yes, but i am not sure what values go where:/
Michele_Laino
  • Michele_Laino
according to @rvc we have: \[\large x = \frac{{\lambda d}}{D} = \frac{{605 \times {{10}^{ - 9}} \times 0.095}}{{85}} = ...meters\]
rvc
  • rvc
85 cm
anonymous
  • anonymous
ok! so we get 6.76 E-10 ?
Michele_Laino
  • Michele_Laino
oops.. \[\large x = \frac{{\lambda d}}{D} = \frac{{605 \times {{10}^{ - 9}} \times 0.095 \times {{10}^{ - 3}}}}{{85 \times {{10}^{ - 2}}}} = ...meters\]
rvc
  • rvc
it should be 0.85 ?
Michele_Laino
  • Michele_Laino
yes you are right! @rvc
rvc
  • rvc
thanks
anonymous
  • anonymous
ok we get 6.76 E -11 ?
anonymous
  • anonymous
i am confused then, what is our solution? :/
Michele_Laino
  • Michele_Laino
ok! so how many mm?
anonymous
  • anonymous
5,400 mm?
Michele_Laino
  • Michele_Laino
you are right! it is not an option of yours @iheartfood
anonymous
  • anonymous
ok! so which would i choose? :/
rvc
  • rvc
the formula
Michele_Laino
  • Michele_Laino
then try this formula: \[\Large \lambda = \frac{{dx}}{D} \to \quad x = \frac{{\lambda D}}{d}\]
rvc
  • rvc
@Michele_Laino the d
anonymous
  • anonymous
please, if you can show me how to use it? i am not sure what to plug in
rvc
  • rvc
yep the formula is wrong :( sorry
Michele_Laino
  • Michele_Laino
we have to change the formula @rvc @iheartfood
anonymous
  • anonymous
ok! how do we do that?
anonymous
  • anonymous
@Michele_Laino ? :)
Michele_Laino
  • Michele_Laino
here is the right step: \[\Large x = \frac{{\lambda D}}{d} = \frac{{605 \times {{10}^{ - 9}} \times 85 \times {{10}^{ - 2}}}}{{0.095 \times {{10}^{ - 3}}}} = ...meters\]
rvc
  • rvc
\(\large x = \frac{{\lambda D}}{d} = \frac{{605 \times {{10}^{ - 9}} \times 85 \times {{10}^{ - 2}}}}{{0.095 \times {{10}^{ - 3}}}} = ...meters\)
anonymous
  • anonymous
ok! and we get 0.005!
Michele_Laino
  • Michele_Laino
I got this: 541315.7*10^(-8) meters
anonymous
  • anonymous
ohh ok oops! so +0.5 is the solution?
rvc
  • rvc
541.31 X 10^{-5}
anonymous
  • anonymous
i still get 0.005...
Michele_Laino
  • Michele_Laino
it is: 54135 * 10^(-5) mm
anonymous
  • anonymous
0.54! okay! so our solution is +0.5!
rvc
  • rvc
my ans was in meters
anonymous
  • anonymous
yay! thanks both!! onto the next!
Michele_Laino
  • Michele_Laino
:)

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