Question

Can you please explain step by step how to solve these type of questions please.

White light is a mixture of all colors with wavelengths between 400 and 700m. If we shine white light on two narrow slits spaced 1 x 10 m apart and place a screen 3 m behind the slits, we will see sn interference pattern a little different from the one we saw in DL. Notice that every wavelenght of light has its n = 0 bright spot in the same location, but the positions of all other points of constructive interference depend on the wavelength (and thus on the color). Therefore, we will see a bright white spot at the center of the pattern, flanked by "rainbows. a), what is the distance on the screen between the white spot at the center of the pattern and the n = 1 constructive interference point for 700 nm light? Please use the approximation that tan θ sin θ for small θ. (Note that this is the "red end of the first rainbow.) we use nA dsin θ for n I, λ 700 mm 7 x 10-7 m, end d 1 x 10-s rn. For sine, observe that the right trialigle formed between the 1 ny,黯= 0 rHy, and the pattern on the wall has D 3 mi as its Bdjacent leg and the distance we Nre looking for (which we shall call Δ2) as the opposite leg. Hence we have tan@ Using the given approximation tan θ ns sine, we get: (3 m)(7 x 10-7 m) x 10- (1)(7 × 10-7 m) = (1 x 10--5 m) 3 nm b), what is the distance on the screen between the white spot at the center of the pattern and the n = 2 constructive interference point for 400 nm light? Again, use the approximation that tan θ ~ sin θ for small θ. (Note that this is the "blue end" of the second rainbow.) The method is the same ing these in, w ge in part a), the only difference is that λ :-400 nm 4 x 10-1 m ind n Plug- (3 Ⅱ)(2)(4 × 10-7 m) 1 x 10-811 0.21 m= 24 cm 3 1m c). Find the smallest integer such that the 400 nm (blue) end of the (n 1)st rainbow is closer to the central bright spot than the 700 nm (red) end of the nth rsinbon Sinoe by our previous results, A ith red and whemever them the (n+1)st blue end is closer to the center of the patterm than the Since n must be an integer, the smallest n that will work is the smallest integer greater than , which is 2