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# HW7 - PHYS 4D Solution to HW 7 Problem Giancoli 35-2(I...

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PHYS 4D Solution to HW 7 February 21, 2011 Problem Giancoli 35-2 (I) Monochromatic light falls on a slit that is 2 . 60 × 10 3 mm wide. If the angle between the first dark fringes on either side of the central maximum is 32 . 0 (dark fringe to dark fringe), what is the wavelength of the light used? Solution: The angle from the central maximum to the first dark fringe is equal to half the width of the central maximum. Using the angle and Eq. 35-1, we calculate the wavelength used θ 1 = 1 2 θ = 1 2 32 . 0 = 16 . 0 , sin θ 1 = λ D λ = 717 nm. Problem Giancoli 35-4 (II) Consider microwaves which are incident perpendicular to a metal plate which has a 1 . 6- cm slit in it. Discuss the angles at which there are diffraction minima for wave-lengths of (a) 0 . 50 cm , (b) 1 . 0 cm , and (c) 3 . 0 cm . Solution: (a) Using Eq. 35-2, m = 1 , 2 , 3 , ... to calculate the possible diffraction minima, when the wavelength is 0 . 50 cm , D sin θ m = θ m = sin 1 ( D ) . θ 1 = 18 . 2 , θ 2 = 38 . 7 , θ 3 = 69 . 6 , θ 4 = no solution. θ 1 , θ 2 and θ 3 give three diffraction minima. (b) Using the new wavelength and repeat the above process, we find θ 1 = 38 . 7 , θ 2 = no solution. θ 1 gives the only diffraction minima. (c) Using the new wavelength and repeat the above process, we find θ 1 = no solution. No diffraction minima. Problem Giancoli 35-6 (II) Monochromatic light of wavelength 633 nm falls on a slit. If the angle between the first bright fringes on either side of the central maximum is 35 , estimate the slit width. . Solution: The angle from the central maximum to the first bright maximum is half the angle between the ﬃrst bright maxima on either side of the central maximum. The angle to the first maximum is about halfway between the angle to the first and second minima. We use Eq. 35-2, setting m = 3 2 , to calculate the slit width, D . θ 1 = 1 2 θ = 1 2 35 . 0

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HW7 - PHYS 4D Solution to HW 7 Problem Giancoli 35-2(I...

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