Light Refection and Snell's Law

[myersb05]

The light beam shown in Figure P22.21 makes an angle of ϕ = 23.5° with the normal line NN' in the linseed oil. Determine the angles θ and θ'. (The refractive index for linseed oil is 1.48.)
The normal line makes a 90 angle with the surface of the linseed oil. The light then passes through water but I need the linseed oil angle to get the water angle.

Snell's Law

So I used 1*sin(23.5)=1.48*sin(theta) and solved for theta. I got 15.63 degrees which is incorrect

 

[Kurdt]

If the angle is in the linseed oil then it is the angle after it has been refracted at the air linseed interface. Thats my guess anyway without seeing the diagram.

 

[Redbelly98]

Snell's Law

So I used 1*sin(23.5)=1.48*sin(theta) and solved for theta. I got 15.63 degrees which is incorrect

That equation works for an air-to-linseed oil interface or a vacuum-to-oil interface.

You have a water-oil interface; the refractive index of water must enter into Snell's law somehow.

edit added:
Plus, use Kurdt's suggestion of putting the angle on the correct side of the equation :-)

 

[Kurdt]

I think Redbelly since its asking for two angles that it will be the linseed oil between air and water. Like I said before however, a diagram would be nice.

 

[Redbelly98]

I agree that showing us the diagram would have been helpful, so that we wouldn't have to spend time discussing and guessing at what is going on.