# Diffraction immersed in water

## Homework Statement

Light of wavelength 676 nm in vacuum is incident on a single slit whose width is 1.60 × 10-5 m. The setup is then immersed in water whose index of refraction is 1.33. What is the angle θ that locates the third dark fringe with respect to the central bright fringe?

## Homework Equations

1. sin$$\alpha$$ = (m +1/2)$$\lambda$$/d
where $$\alpha$$ is the angle of diffraction, $$\lambda$$ is the wavelength, and d is the width of the slit
2. sin$$\angle$$i / sin$$\angle$$r = n
where i is the angle of incidence, r is the angle of refraction, and n is the refractive index

## The Attempt at a Solution

sin$$\alpha$$ = (3 + 1/2)(676)(10^-9)/(1.6)/(10^-5)
sin$$\alpha$$ = .147875
$$\alpha$$ = 8.5037997

sin$$\alpha$$ = sin$$\angle$$i ( i presume, so...)
.147875/sin$$\angle$$r = 1.33
sin$$\angle$$r = .11118421
$$\angle$$r = 6.383584609

i'm not sure is theta = $$\angle$$r

## Answers and Replies

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i figured it out. i was using the wrong formula for diffraction. instead of m + 1/2, i should have been using m

sin(theta) = m(lambda)/(d) instead of m+1/2