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Homework Help: Reviewing for final, please help

  1. May 8, 2004 #1
    I'm reviewing for a final in my physics class and was wondering if someone could help me with the problems I got wrong (maybe show the equation and solution or something). Thanks!

    1) Light has a wavelength of 521.5 nm and a frequency of 4.690 x 1014 Hz when traveling through a certain substance. What is the refraction index of this medium?

    2) A flat sheet of ice (n = 1.309) has a thickness of 2.4 cm. It is on top of a flat sheet of crystalline quartz (n = 1.544) that has a thickness of 1.2 cm. Light strikes the ice perpendicularly and travels through it and then through the quartz. In the time it takes the light to travel through the two sheets, how far (in cm) would it have traveled in a vacuum?

    3) The drawing shows a rectangular block of glass (n = 1.52) surrounded by a liquid with n = 1.65. A ray of light is incident of the glass at point A with a 30.0° angle of incidence. At what angle does the ray leave the glass at point B?

    4) What is the critical angle for light emerging from a transparent medium into air, if the refraction index of the medium is 1.540?

    5) A converging lens has a focal length of 85 cm. A 14-cm-tall object is located 121 cm in front of this lens. (a) What is the image distance? (b) What is the image height? Be sure to include the proper algebraic sign.

    6) An object is located 38.2 cm to the left of a converging lens whose focal length is 59.2 cm. (a) Determine the image distance and (b) the magnification.

    7) An object is 27 cm in front of a diverging lens that has a focal length of -14 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 2.1?

    8) An object is located 8 cm in front of a converging lens (f = 4.1 cm). Determine where the image is located.

    9) When a diverging lens is held 15 cm above a line of print, the image is 8.1 cm beneath the lens. What is the focal length of the lens?
  2. jcsd
  3. May 8, 2004 #2


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    Some hints...

    1 & 2. How is the refractive index defined ? Are there terms in the definition that can be calculated from the given data ? If there is more than one definition, use the relevant one.

    3. Use the other definition of refractive index to figure out angles.

    4. How does light get deflected when traveling from a denser to a rarer medium ? What happens to the refracted ray as you continuously increase the incident angle ? Look at the definition of critical angle.

    5 to 9. These are direct formula substitution problems. There's just one fomula to use for them all. Just be careful with the sign conventions.
  4. May 8, 2004 #3


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    Find the velocity of the light.

    [tex]V = \lambda f[/tex]

    Now use the index of refraction formula.

    [tex]n = \frac{c}{V}[/tex]

    Same formula as before.

    [tex]V = \frac{c}{n}[/tex]

    Draw a picture.

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

    Same formula as above but you have assumed stuff going on. Remember that for the critical angle, the n value for air is what it always is (around 1 I think), and theta for air is 90 degrees. So solve for this
    1.54 sin(theta) = 1 sin(90)

    I don't remember how to do lens questions.
  5. May 8, 2004 #4
    For the first problem, how do I find n=C/V? I know v=2.45x10^17 but how do I figure out C?
  6. May 8, 2004 #5


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    Did you actually show up for any of the classes? :biggrin:
    c is the speed of light.
  7. May 8, 2004 #6


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    You mean 107 not 1017, right? c is the speed of light in a vacuum and is a constant, approximately 3*108 m/s

    Edit: A little slow to reply as usual; didn't mean to butt in ShawnD.
  8. May 8, 2004 #7
    Hmmm here's what I've done: 3.00x10^8/2.45x10^7=12.244 but the program is saying the answer is incorrect. What did I do wrong?
  9. May 8, 2004 #8
    You should check your speed again. V = Wavelength x frequency.

    If you do it carefully, you should get an Index of refraction of 1.22.
  10. May 8, 2004 #9
    Ok, here's what I did for problem 3. I thought this answer was correct but it isn't. What did I do wrong?

    sin 30 deg(1.65)=sin theta g (1.52)
    sin -1[sin 30 deg(1.65)/(1.52) = theta g
    = 32.87 deg
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