Interference in thin films

Discussion in 'Physics' started by logearav, Oct 6, 2011.

  1. logearav

    Thread Starter Member

    Aug 19, 2011
    Members, kindly see my attachment.
    A ray of monochromatic light AB is incident on the surface K of film. It is partly reflected along BC and partly refracted into the film along BD. At the point D on the surface K', the ray of light is partly reflected along DE and partly transmitted out of the film along DG. The reflected light then emerges into air along EF which is parallel to BC.
    In my attachment ZX is the normal drawn. So,<ABZ is angle of incidence.
    <ZBC----angle of reflection;
    Now my question is how <ABZ and <MEB are equal? Both are designated as 'i' in the attachment. Because, angle between incident ray and normal drawn qualifies to be angle of incidence. There is no incident ray in the triangle MEB.
    Please help, revered members
  2. Georacer


    Nov 25, 2009
    How about this:
    \hat{ABZ}=\hat{ZBM}<br />
=90^o-\hat{MBE}<br />
=90^o-(180^o-\hat{BME}-\hat{MEB})<br />
    logearav likes this.
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    Another way, <ABZ = <ZBM (by angle of incidence = angle of reflection)

    As Z-X is a perpendicular, <ZBE = 90 (right angle)

    Therefor: <MBE = 90 - <ZBM = 90 - <ABZ

    For right triangle BME:
    <MEB = 90 - <MBE = 90 - (90 - <ABZ)

    <MEB = <ABZ
    logearav likes this.
  4. logearav

    Thread Starter Member

    Aug 19, 2011
    Thanks for the replies, Georacer and ErnieM. As always, this is a forum which always helps to learn concepts in a better way. I feel proud in being a member of this group.