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Posted by on Thursday, January 24, 2013 at 3:22pm.

What is the angle of refraction of a light ray that enters the surface of a calm lake from air at an angle of 45o to the normal? The refractive index of water is 1.333.


I don't even know how to start this or what the formula is. Thanks for your help in advance. It really is appreciated.

  • Physics - , Thursday, January 24, 2013 at 3:37pm

    Snell's law:

    n1 Sin(theta)=n2 Sin(thea)

    Index of refraction of air (n1)=1.00
    angle of incidence is 45 to the normal (sin theta)

    Index of refraction of lake (n2)=1.33

    solve for theta

  • Physics - , Thursday, January 24, 2013 at 3:45pm

    Okay so then 1(sin45) = 1.33(sinx)

    I got 32.04degrees?

  • Physics - , Thursday, January 24, 2013 at 3:49pm

    n1 Sin(theta)=n2 Sin(theta)

    Index of refraction of air (n1)=1.00
    angle of incidence is 45 to the normal (sin theta)

    Index of refraction of lake (n2)=1.33

    solve for theta

    (1.00/1.33)sin45=sin(theta)

    0.544=sin(theta)

    sin-1 (0.544)=theta

    33.0 degrees=theta

  • Physics - , Thursday, January 24, 2013 at 3:55pm

    I got 32.1 degrees. The problem worked out should read

    0.532=sin(theta)

    sin-1 (0.532)=theta

    32.1 degrees=theta

  • Physics - , Thursday, January 24, 2013 at 4:22pm

    Thank you so much

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