1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I really need someone to help me with these physics math problems

  1. Apr 24, 2005 #1
    Since no one seems to be helping me, maybe you will help me when I post my exact work that I did. I just need someone to check them all and make sure I did the right equations and if I got the right answers or not. This is for a grade, and I desperately need a passing grade on this worksheet to even pass the class. I made an honest attempt at each.

    10. the picture that we see in a mirror is called the
    my answer=image
    11. the actual object that makes a picture that we see on a mirror is called the
    my answer=specular

    -problems- show your work

    15. What is the wavelength of a wave with a frequency of 100.4 MHz?

    f=100.4 MHz



    =.335 m

    16. A substance allows light to travel through it with a velocity of 2.45 x10^8 m/s. What is the substances index of refraction?

    v=2.45x10^8 m/s


    2.45 x 10^8
    3 x 10^8


    17. What is the intensity of a light that has a lumnious flux of 750 lumens?

    P= 750 lm
    E= ?


    E=750 lm/4(pie)

    E=589.05 cd

    18. What is the illuminance provided by a light with a luminous flux of 1120 lm when it is 2.3 meters away?

    E= ?
    P= 1120 lm
    d= 2m


    E=1120 lm/4(pie)2^2

    E=22.28 lx

    19. What would the angle of refraction be if a light ray starts in a diamond and goes into the air it has an angle of incidence of 9 degrees.

    90 degrees



    =24.44 degrees

    20. What is the critical angle of light going from quartz to air?


    sin x= 1/1.544 = .647


    =40.37 degrees

    22. What is the illuminance of a light when it is 4.6 m away from a light with an intensity of 78 candle power?

    d= 4.6 m
    cd= 78
    P= 980.2




    E=3.69 lx
  2. jcsd
  3. Apr 24, 2005 #2
    I post my work and I still get no help. I have asked politely and everyone refuses to help me. I don't get it.
  4. Apr 24, 2005 #3
    I would help if I could... no one seems to ever reply to my questions either. One time I even went as far as scanning my paper with what I had done so far and no one replied.
  5. Apr 24, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Check the arithmetics in #17.

  6. Apr 24, 2005 #5
    What'd you see wrong in 17?
  7. Apr 24, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    Isn't one candela one lumen in one sterradian...?

  8. Apr 24, 2005 #7
    I don't think so
  9. Apr 25, 2005 #8


    User Avatar
    Science Advisor
    Homework Helper

  10. Apr 25, 2005 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The answer to 10 sounds reasonable enough. But your answer to 11 doesn't make a lot of sense to me. As far as I know, "specular" is a description, not a thing. It describes reflection from a smooth, polished surface, if I remember right. That having been said, I have no idea what #11 could be asking for, especially since they already used the word "object".

    15. [tex] c = f \lambda [/tex]. Solve for lamda. I don't your answer is correct. Recheck your calculations and algebra.

    16. Your formula for the index of refraction is correct. *Note that you have reversed the numerator and denominator after plugging the numbers in*. Nevertheless, I'm assuming you did it correctly on your worksheet, since your answer is > 1.

    Sorry...I don't remember enought about lumens, candela, lux and all those other crazy units to help you there.

    Note: [itex] \pi [/itex] is spelled, "pi".

    I think you've got the basic idea, *except that we are going from diamond to air, not the other way around!*. So let air be medium 2, and diamond be medium 1. Applying Snell's law, it should be:

    [tex] n_1 \sin \theta_1 = n_2 \sin \theta_2 [/tex]

    n2 = 1 for air.
    n1 = ? (I'm assuming 2.417, since that is what you wrote)

    [tex] \theta_2 = \sin^{-1} (n_1 \sin(9^o)) [/tex]

    I think you did this one correctly. Key concept: at the critical angle of incidence, the angle of refraction is 90 degrees. The sine of 90 degrees is 1.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?