Question

Light of wavelength 459 nm falls on two slits spaced 0.280 mm apart. What is the required distance from the slit to a screen if the spacing between the first and second dark fringes is to be 4.00 mm? m MY NOTES PRACTICE ANOTHER 5. [-/10 Points) DETAILS SERCP7 24.P.007 Two radio antennas separated by d - 277 m, as shown in Figure P24.7, simultaneously transmit identical signals of the same wavelength. A radio in a car traveling due north receives the signals. 400 m 900 m 1000 m Figure P24.7 (a) If the car is at the position of the fourth maximum, what is the wavelength of the signals? (b) How much farther must the car travel to encounter the next minimum in reception? (Hinti Determine the path difference between the two signals at the two locations of the car. m

Answer 1

Position Dark Fringes on of Screen is, Youne - (m+ 1 m+ ID d where n= 0, 1, 2, 3, The Spaing between the first & second dark fringer, Ayo (1+2) X-1001) if us 3 D= (ydd a) Da (4 x 1079 m) (0.28 x 103 m) 459 X 10.9 m 2.44 m Distance Slit is! og Screen from 2.44 m. Ans.

(a the Car is at position of 4th maximum + So, At m= 4 maximum, tan o = 4oom 1000 m 0,400 o 21.8° = So, x = dsino = (277m) sin(21.8°) 4 À = 25,72m (6) The next minimum encountered is the m = 4 minimum & at that point, dsino = (m mt I 2 dsino 올 = sino = 9 x 2 2 d to 3 sino= 9 25.72 2 277 225.72

dsino = 9 x = sino 9 X 2 d 2 sino = 0.417 old 25.72m 277m O = 24.65° (1000 m). tau (24.6 y = 457.8m 6°) = 8. The Car must travel Additional [57.8m So, Ans.

* Answers:

• D = 2.44 m
• (a) λ= 25.72 m , (b) 57.8 m