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-   -   Physical optics, intensity and power (https://www.askmehelpdesk.com/showthread.php?t=325978)

  • Mar 7, 2009, 01:48 AM
    Dina Saeed
    Physical optics, intensity and power
    Dear Mr.

    Thanks for the great site. I've a question in Physical optics for pre-pharmacy students.
    My question is, sunlight strikes a solar panel communication satellite at an angle of 65 degrees and at a power of 3200 watts, what's the power and intensity when the angle becomes 35 degrees instead of 65?

    Regards,
    Dina A. Saeed
  • Mar 10, 2009, 08:45 AM
    ebaines

    The power of the incident solar radiation is proportional to th apparent size of the array as "seen" by the sun. This apparent size is A*cos(angle), where A is the array's true size and "angle" is the angle of incidence to the sun. I am assuming here that the nagle is measured from the perpendicular axis of the array, so that if the angle is 0 degrees the solar array is full face on towards the sun. So the ratio of power given the two conditions are:

    x watts/3200 watts = cos(35 degrees)/cos(65 degrees).

    Solve for x.

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