Recent content by SEEDS

  1. S

    Calculating the Distance Between Interference Maxima: Yellow Light

    Homework Statement a)Yellow light Lambda= 589nm falls on a double slit. The distance between two slits is d=0.5 mm the width of each slit is a=100 micrometer. A screen showing the interference pattern is located 4 m behind the slits. Homework Equations What is the distance between...
  2. S

    D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

    ok let's check it fn = n v/2L ---A fn+1 = (n+1) v/2L ----B fn = 466.2 Hz , fn+1 = 582.7 Hz Therefore A becomes 466.2 = n (343 / 2L ) 2.718L = n ------(i) and B becomes by fn+1 582.7L = 171.5 n + 171.5 ---(ii) putting n from i L = 1.471 meter and n is approx = 4...
  3. S

    D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

    f_n = n\frac{v}{2L} I am applying it like n = (fn * 2 L) / v but n1=2.718 L n2=3.397 L n3=4.076 Lstill confusion how to find L to proceed finally for the harmonics ?
  4. S

    D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

    The fundamental frequ. of an open pipe is f0 = v/(2*L1)
  5. S

    D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

    If I am using fn = n f1 so i am getting fractional figures 466.2 / 264 = n and i guess n should'nt be fractional . 582.7 / 264 = n 699.2 / 264 = n
  6. S

    D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

    could you confirm me about the fundamental fr in (C)
  7. S

    D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

    Ohh Sorry ! For C Please take a look at my first post.
  8. S

    D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

    Homework Statement One organ pipe open on both ends and one organ pipe open at one end and closed at the other are tuned on the same fundamental tone with frequency f = 264 Hz. a)How long is each pipe? b) Determine the first three harmonics of each pipe. c) There is an organ pipe open at both...
  9. S

    Black Body Radiation: Find Temperature of Panel PM in Radiation Equilibrium

    Ohh Sorry .. Now i got it , Tm4 = (T24 + T14)/2 Tm = 334.21 Kelvin How does the radiation balance of P2 change with the insertion of PM? What i m getting is radiation balance of P2 with P1 will change after the insertion but how can we define it ?
  10. S

    Black Body Radiation: Find Temperature of Panel PM in Radiation Equilibrium

    Thanks a lot for the help , sA(TM4 - T14) = sA (T24 - TM4) By canceling sA (TM4 - T14) = (T24 - TM4) Tm4 = (T24 - T14)/2 Tm4 = 6.9106 10^9 . How does the radiation balance of P2 change with the insertion of PM? Thanks again
  11. S

    Black Body Radiation: Find Temperature of Panel PM in Radiation Equilibrium

    Ok, Thanks a lot for the hint . I tried it please take a look at it ... According to Stefans LAW P=SAT^(4) P = Power radiated in W (J/s) s = Stefan's Constant 5.67 x 10-8 W m-2 K-4 A = Surface area of body (m²) T = Temperature of body (K)To Find out the temperature of Pm :- P...
  12. S

    Black Body Radiation: Find Temperature of Panel PM in Radiation Equilibrium

    ***** SHEETS two ideally black of the area 1:5 mETER SQUARE are located opposite of each other in vacuum. The first SHEET P1 has the FIXED temperature T1 = 0o C, the second SHEET P2 has the fIXED temperature T2 = 100o C. Now an additional black panel PM of the same size is brought between the...
Back
Top