Question

Light of wavelength 650 nm falls on a slit that is 3.90×10⁻³ mm wide.

How far the first bright diffraction fringe is from the strong central maximum if the screen is 13.0 m away.

This is my last attempt; I've put 3.25 m, 2.20 m, 2.17 m

Answer 1

Wavelength of light = 1 = 650 nm Inm=10 9 m » = 650 X 10 m = 6-58107m width of slit = Q = 3.90 X10-3 mm : I mm = 10-3 m ) a = 3.90 X 100 m distance between slit and screen =D = 13-om . A first bright fringe amejo CTO : - central maxima Slit C D I screen where o, is angle of first bright fringe y, is distance of first bright fringe from central maxima. Now, relation between angle of nth maxima (bright) fringe and warelength is given by a sin on = (2n+]) ; n = 1,2,3,--... For first bright fringe n=1 =) a sin o, = (2x1 +1) =) a sino, = 3 =) sino, = 3 4 » sino = 3 X 6.5 X10-7m - 2 x 3.goxloom = sino = 1958 loom - 7.8 xlo-om o sino, = 0.25 =) 0, = sin(0.25) => 0, = 14.4775° From right angle A SAO , 1 tan o, = AO - tano, = 21 da

- y, = tan Q = y = (13.0m) tan (14.47750) =) Jy = (13.0 m) 10.2582) => Y, = 3.3566 m Ž 3.36 m So, first bright diffraction fringe will be at 3.36 m - from central maximum.