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    Practice problems in Optics

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    1. Light with a wavelength of 486 nm in air is passed through glass with an index of refraction of 1.54. (a) What is the wavelength of the light in the glass? What is the frequency of the light (b) in air and (c) in the glass?

    2. Using Huygens' principle (rather than Snell's law), find the angle of refraction when light is incident on the interface between air (n = 1.00) and glass (n = 1.50) at an angle of 45 degrees.

    3. Red light of wavelength 632.8 nm is incident on a pair of slits whose separation is 0.75mm. What is the separation of neighboring interference maxima on a screen 180cm from the slits?

    4. How thick should a sheet of mica (n=1.58) be if it is to be as thin as possible and still give rise to destructive interference for reflection, given that the incident light is in the red (650 nm) part of the spectrum?

    5. The index of refraction of ice (n=1.31) is slightly less than that of water (n=1.33). Is it theoretically possible to have a sheet of ice floating on water that does not reflect the green (550nm) part of the spectrum from the sun when it is directly overhead? If so, what is the minimum thickness of the sheet?

    6. A single-slit diffraction pattern is formed when light of 633nm is passed through a narrow slit. The pattern is viewed on a screen placed one meter from the slit. What is the width of the slit if the width of the central maximum is 2.53cm?

    7. Laser light 633nm is normally incident on a 0.100-mm-wide slit in a thin opaque plastic sheet floating on the surface of a swimming pool. Calculate the width of the central maximum at the bottom of the pool 4.50 m below the surface?

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    Solution Preview

    I have given references to almost all the problems. Please look at those sites for detailed explanations.

    1. Refractive index = lambda_air/lambda_glass = 1.54 (given)

    ==> Lambda_glass = lambda_air/1.54 = 3.1558*10^-7 = 315.58 nm --Ans

    Frequency remains unaltered in new medium.
    Frequency = c/lambda_air = 3*10^8/486*10^-9 = 6.173*10^14 Hz --Ans

    2. Sin(t1) = [c/v] * Sin(t2) where v is the velocity of the wave

    t1 = 45, but c/v = the refractive index of the medium = 1.5
    Thus t2 = arcsin [v*Sin(t1)/c]
    = arcsin [Sin(45)/1.5] = 28.13degree --Ans

    Please see this page for a good explanation

    3. Separation between fringes is given by the formula,
    y = ...

    Solution Summary

    A set of seven practice problems in optics involving diffraction, reflection and refraction.