Is My Calculation of Light Refraction and Wavelength in Water Correct?

[Bri]

Could someone tell me if I did this right?
A narrow beam of sodium yellow light, with wavelength 589 nm in vacuum, is incident from air onto a smooth water surface at an angle of incidence of 35.0 degrees. Determine the angle of refraction and the wavelength of the light in water.
I did...
sin(35 degrees)=1.333*sin(theta)
theta = 25.486 degrees

589 nm = 1.333*lambda
lambda = 441.86

I'm just not sure I used the right equation/setup.

And I can't figure out this problem...
Unpolarized light in vacuum is incident onto a sheet of glass with index of refraction n. The reflected and refracted rays are perpendicular to each other. Find the angle of incidence.
I figured Theta1 + Theta2 = 90 degrees
and I tried to find a way to solve it using sin(Theta1) = n*sin(Theta2) but it's not working out.

Thanks.

 

[vsage]

The first part looks right to me I guess. Think about \Theta_2 = 90 - \Theta_1 and plug it into your formula. Remember that sin(90-x) = cos(x)

 

[Bri]

The first part looks right to me I guess. Think about \Theta_2 = 90 - \Theta_1 and plug it into your formula. Remember that sin(90-x) = cos(x)

I've tried it that way, I wasn't able to get anywhere with it.

 

[vsage]

I don't really see a definitive answer for part 2, but since sin(\theta_1) = n\times cos(\theta_1), \theta_1 = tan^{-1}(n)