- #1
mrmonkah
- 24
- 0
1. Light propagating from water into air is incident on the water surface at the angle of total internal reflection. (a-int = 48.75 degrees) A plane parallel glass plate (n-glass = 1.5) is brought into contact with the surface of the water. Calculate the angle 'alpha-external' (a-ext for notation simplicity) between the top surface of the glass plate and the direction at which the light will emerge from the glass plate.
n-air = 1
n-water = 1.33
n-glass = 1.5
2. (n-water)(sin(Theta-water)) = (n-glass)(sin(theta-glass)), (n-glass)(sin(theta-glass)) = (n-air)(sin(theta-air))
3. I have combined the two equations to get:
sin(theta-air) = (n-air/n-water)sin(a-int)
where by not needing to take into account the refractive index n-glass. this gives me a value of:
(theta-air) = 48.75, so 90-(theta-air) = a-ext = 41.25
I have then made the calculations separately for each barrier and came out with a value of 1.1 for a-ext? Could anyone verify if either of these solutions are correct?
The question then goes on to ask what happens if (n-glass) = 1.2, 1.4 and 2. So i presume i must take into consideration the glass block overall, and my first attempt at combining the 2 snell equations is incorrect.
n-air = 1
n-water = 1.33
n-glass = 1.5
2. (n-water)(sin(Theta-water)) = (n-glass)(sin(theta-glass)), (n-glass)(sin(theta-glass)) = (n-air)(sin(theta-air))
3. I have combined the two equations to get:
sin(theta-air) = (n-air/n-water)sin(a-int)
where by not needing to take into account the refractive index n-glass. this gives me a value of:
(theta-air) = 48.75, so 90-(theta-air) = a-ext = 41.25
I have then made the calculations separately for each barrier and came out with a value of 1.1 for a-ext? Could anyone verify if either of these solutions are correct?
The question then goes on to ask what happens if (n-glass) = 1.2, 1.4 and 2. So i presume i must take into consideration the glass block overall, and my first attempt at combining the 2 snell equations is incorrect.