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Homework Help: 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.


    analysis:
    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?

    Given:
    f=100.4 MHz
    3x10^8

    f=100.4x10^6

    1004000000
    -----------
    3x10^8

    =.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?

    Given:
    v=2.45x10^8 m/s
    3x10^8

    n=c/v

    2.45 x 10^8
    ------------
    3 x 10^8

    =1.22

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

    Given:
    P= 750 lm
    E= ?

    E=P/4(pie)

    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?

    Given:
    E= ?
    P= 1120 lm
    d= 2m

    E=P/4(pie)d^2

    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.

    Given:
    airn=1
    n=?
    90 degrees

    n1sinQ1=n2sinQ2

    sin(9)=n2sin(nr)
    2.417xsinQ2=1sin90

    =24.44 degrees

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

    Given:
    nquartz=1.544

    sin x= 1/1.544 = .647

    sin-1(.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=P/4(pie)d2
    cd=P/4(pie)

    CD4(pie)=P
    78(4)(pie)

    E=980.2/4(pie)(4.6)^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

    dextercioby

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    Check the arithmetics in #17.

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

    dextercioby

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

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

    dextercioby

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  10. Apr 25, 2005 #9

    cepheid

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    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.
     
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