Sunlight is used in a double-slit interference experiment. Thefourth-order maxim

Question

Sunlight is used in a double-slit interference experiment. Thefourth-order maximum for a wavelength of 470 nm occurs at an angleof = 90°. Thus, it is on the verge of beingeliminated from the pattern because cannot exceed90° in Eq. 35-14. (a) What least wavelength inthe visible range (400 nm to 700 nm) are not present in thethird-order maxima? To eliminate all of the visible light in thefourth-order maximum, (b) should the slitseparation be increased (show 1) or decreased (show 0) and(c) what least change in separation is needed?

Explanation / Answer

For the fourth order maxima m = 4    wave length = 470nm   , =900     since d sin = m /sin               d = 4*470nm/ sin90    If ( A ) m = 3 then 1 = d sin /m             = 4*470 / 3 = 626nm Any wavelength greater than this will not be seen .Thus 600nm < 700nm are absent so then the visible range will be    400nm < < 600nm B) since the slit seperation d is related to as d is directly proportional to .' ' decreases only when d decreases C) since d1 =m1 1 / sin and  d2 = m2 2 / sin d1 ~ d2 = 1/ sin 90 ( 4*400 -3*600 ) = 200nm                                                              

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