# monochromatic light of wavelength 400 mm and 560 mm are incidents simultaneously and normally on double slit apparatus​ whose slit separation  is 0.1 mm and screen distance is 1 m.distance between areas of total darkness will be

Let nth minima of 400nm coincides with m th minima of 560nm, then FORMULA : (2n-1)λ/2 (2n-1)(400/2) = (2m-1)560/2 2n-1/2m-1=7/5 10n-14m+2=0 By inspection m=3,n=4 and m =8, n=11 i.e, 4th minima of 400 nm coincides with 3rd minima of 560 nm. Location of this minima is, FORMULA : Y = {(2n-1)D*λ}/2d Y1 = (2 x 4 -1)(1000)(400 X 10-6)2 X 0.4 = 14 mm. Next 11 th minima of 400 nm will coincide with 8th minima of 560nm Y2 = (2 X 11 - 1)(1000) (560 X 10 -6)2 X 0.1 = 42 mm Required distance is Y2 - Y1 = 28 mm.
• 3
(2n+1)λ1=(2m+1)λ2 2n+12m+1=560400=75 10 n = 14m + 2 By inspection, for m = 2 ; n = 3 m = 7; n = 10 Δs=λ1D/2d[(2n2+1)−(2n1+1)]=28mm.
• 1
Let nth minima of 400nm coincides with m th minima of 560nm, then FORMULA : (2n-1)?/2 (2n-1)(400/2) = (2m-1)560/2 2n-1/2m-1=7/5 10n-14m+2=0 By inspection m=3,n=4 and m =8, n=11 i.e, 4th minima of 400 nm coincides with 3rd minima of 560 nm. Location of this minima is, FORMULA : Y = {(2n-1)D*?}/2d Y1 = (2 x 4 -1)(1000)(400 X 10-6)2 X 0.4 = 14 mm. Next 11 th minima of 400 nm will coincide with 8th minima of 560nm Y2 = (2 X 11 - 1)(1000) (560 X 10 -6)2 X 0.1 = 42 mm Required distance is Y2 - Y1 = 28 mm.
• 1
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The passage contains a physics problem involving the calculation of the distance between areas of total darkness in a double-slit experiment with light of two different wavelengths. The text provides a detailed solution to the problem, including the necessary formulas and steps to find the answer.
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# monochromatic light of wavelength 400 mm and 560 mm are incidents simultaneously and normally on double slit apparatus​ whose slit separation  is 0.1 mm and screen distance is 1 m.distance between areas of total darkness will be

Let nth minima of 400nm coincides with m th minima of 560nm, then FORMULA : (2n-1)λ/2 (2n-1)(400/2) = (2m-1)560/2 2n-1/2m-1=7/5 10n-14m+2=0 By inspection m=3,n=4 and m =8, n=11 i.e, 4th minima of 400 nm coincides with 3rd minima of 560 nm. Location of this minima is, FORMULA : Y = {(2n-1)D*λ}/2d Y1 = (2 x 4 -1)(1000)(400 X 10-6)2 X 0.4 = 14 mm. Next 11 th minima of 400 nm will coincide with 8th minima of 560nm Y2 = (2 X 11 - 1)(1000) (560 X 10 -6)2 X 0.1 = 42 mm Required distance is Y2 - Y1 = 28 mm.
• 3
(2n+1)λ1=(2m+1)λ2 2n+12m+1=560400=75 10 n = 14m + 2 By inspection, for m = 2 ; n = 3 m = 7; n = 10 Δs=λ1D/2d[(2n2+1)−(2n1+1)]=28mm.
• 1
Let nth minima of 400nm coincides with m th minima of 560nm, then FORMULA : (2n-1)?/2 (2n-1)(400/2) = (2m-1)560/2 2n-1/2m-1=7/5 10n-14m+2=0 By inspection m=3,n=4 and m =8, n=11 i.e, 4th minima of 400 nm coincides with 3rd minima of 560 nm. Location of this minima is, FORMULA : Y = {(2n-1)D*?}/2d Y1 = (2 x 4 -1)(1000)(400 X 10-6)2 X 0.4 = 14 mm. Next 11 th minima of 400 nm will coincide with 8th minima of 560nm Y2 = (2 X 11 - 1)(1000) (560 X 10 -6)2 X 0.1 = 42 mm Required distance is Y2 - Y1 = 28 mm.
• 1
What are you looking for?
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The passage contains a physics problem involving the calculation of the distance between areas of total darkness in a double-slit experiment with light of two different wavelengths. The text provides a detailed solution to the problem, including the necessary formulas and steps to find the answer.

TYPE: SINGLE_QUESTION_WITH_ANSWER