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Homework Help: Refractive index snell's law help

  1. Feb 26, 2009 #1
    refractive index... snell's law... help!!


    im having a LOT of trouble with what seems like a basic question. i have been stressing over this for nearly 3 hours now. can anyone help me out? here it is:

    Light propagating from water into air is incident on the water surface at the angle of total internal reflection. A plane parallel glass plate is brought into contact with the surface of water.

    Will any light be observed as a result of that? Justify your answer by using Snell’s law.

    snell's law: n1sin [tex]\theta[/tex]1 = n2sin[tex]\theta[/tex]2

    where n is the refractive index of the medium either side of the boundary, and theta is the angle made by the ray with the normal to the boundary either side of the boundary.

    all i have managed to work out is that the light ray coming in at the reflection angle for water - air boundary would not be able to experience total internal reflection at a water - glass boundary, since the refractive index of glass is greater than water.

    when the ray travels through the glass and meets the glass - air boundary, total internal reflection is possible since air has a lower refractive index than glass.

    but, i am stuck here. i cant find a way to show whether the ray will be totally internally reflected or not. has anyone got any suggestions?

  2. jcsd
  3. Feb 26, 2009 #2
    Re: refractive index... snell's law... help!!

    PS. if you haven't answered because it looks like i haven't tried hard enough or haven't given enough info then please say so, because half the time i ask for help on this site i don't even get a reply even though the question gets viewed. i want to know why.
  4. Feb 26, 2009 #3


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    Re: refractive index... snell's law... help!!

    you need to show some working... to both show you;ve attempted teh problem & it makes it easier for people to see what you're trying to do & how to correct you

    but this how i would look at it

    Water to Air
    nw.sin(tw) = na.sin(ta) = na.1 as at TIR so ta = 90, sin(ta) = 90

    so sin(tw) = na/nw
    (note na is close to 1 but we don't have to cancel so leave for now, might gain some extra insight later)

    what happens when the glass is inserted, tw is the same
    nw.sin(tw) = ng.sin(tg)

    so we have (tg) now write the equation for the glass air interface compare them ..
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