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physics questions please help!?

with 2 comments

1) The table lists the range of wavelengths in vacuum corresponding to a given color. If one looks through a film that has a refractive index of 1.333 and thickness of 340 nm, which color will be 100% transmitted through the film?
2) The table lists the range of wavelengths in vacuum corresponding to a given color. Which one of these colors will produce a diffraction pattern with the widest central maximum, assuming all other factors are equal?
Colors:
red 780-622
orange 622-597
yellow 597-577
green 577-492
blue 492-455
violet 455-390

I don’t really understand how I would figure these out. SO if you could please explain to me how you got your answer and stuff that would be great thanks!


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    Written by admin

    August 18th, 2011 at 12:51 pm

    2 Responses to 'physics questions please help!?'

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    1. Problem 1
      ========
      According to snells law, n_water/nair = w_air/w_water
      1.33 / 1 = w_air/340 nm
      w_air = 1.33 * 340 nm
      w_air = 452 nm which is just in the violet range.

      Problem 2
      ========
      The diffraction grating formula is
      n*wavelength = d * sin(x) Sin (x) is what you are interested in.
      d is a constant for the grating. and n tells you which fringe you are measuring. So take n = 1 in all cases. All that varies is the wavelength and Sin(x). Sin(x) describes how large the central fringe area is. The larger the wavelength, the larger the fringe area. That would made 680 nm with the largest central area which is red.

      jcherry_99

      18 Aug 11 at 12:51 pm

    2. I can’t do number 2

      Number one needs the two reflected rays to have a path difference of 1.5 or 2.5 wavelengths

      2 . t . µ + ( 0.5 . wavelength ) = path difference = 1.5 or 2.5 wavelength

      For maximum transmission

      wavelength = 2 . 340 . 10^–9 . 1333 = 906 nm ( outside the range )

      or 340 . 10^–9 . 1.333 = 453 nm

      The colour 100% transmitted will be violet.

      rex

      18 Aug 11 at 12:51 pm

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