Vblue-1.224x108 m/s ± A Sparkling Diamond I Submit My Answers Give Up A beam of

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Vblue-1.224x108 m/s ± A Sparkling Diamond I Submit My Answers Give Up A beam of white light is incident on the surface of a diamond at an angle a .(Figure 1) Since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. The indices of refraction in diamond are nred-2.410 for red light and nble2.450 for blue light. The surrounding air has nair = 1.000. Note that the angles in the figure are not to scale Correct Part C Derive a formula for o, the angle between the red and blue refracted rays in the diamond Express the angle in terms of nred, Mblue, and 0a. Use ni1. Remember that the proper way to enter the inverse sine of in this case is asin(x) Hints Figure 1 of 1 = 1.285 Submit My Answers Give Up red Incorrect; Try Again; 5 attempts remaining lue Part D This question will be shown after you complete previous question(s). Continue

Explanation / Answer

Apply snell's law

nair * sin(1) = ndiamond * sin(2)

For red light, the above becomes:

sin(1) = nred * sin(2, red)
2, red = sin^-1(sin(1) / nred)

And for blue light:

sin(1) = nblue * sin(2, blue)
2, blue = sin^-1(sin(1) / nblue)

Since we want the difference between the angles:

= sin^-1(sin(1) / nred) – sin^-1(sin(1) / nblue)

= sin^-1(sin(1) / nred) – sin^-1(sin(1) / nblue)

= sin^-1 ( sin(45)/2.410) - sin^-1 ( sin 45/2.450)

=0.287 degree

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