A beam of light travels from water into a piece of diamond in the shape of a triangle, as shown in the diagram. Step-by-step, follow the beam until it emerges from the piece of diamond
(a) How fast is the light traveling inside the piece of diamond?
The speed can be calculated from the index of refraction:
(b) What is , the angle between the normal and the beam of light inside the diamond at the water-diamond interface?
A diagram helps for this. In fact, let's look at the complete diagram of the whole path, and use this for the rest of the questions. The angle we need can be found from Snell's law:
(c) The beam travels up to the air-diamond interface. What is , the angle between the normal and the beam of light inside the diamond at the air-diamond interface?
This is found using a bit of geometry. All you need to know is that the sum of the three angles inside a triangle is 180°. If is 24.9°, this means that the third angle in that triangle must be 25.1°. So:
(d) What is the critical angle for the diamond-air interface?
The speed can be calculated from the index of refraction:
(b) What is , the angle between the normal and the beam of light inside the diamond at the water-diamond interface?
A diagram helps for this. In fact, let's look at the complete diagram of the whole path, and use this for the rest of the questions. The angle we need can be found from Snell's law:
(c) The beam travels up to the air-diamond interface. What is , the angle between the normal and the beam of light inside the diamond at the air-diamond interface?
This is found using a bit of geometry. All you need to know is that the sum of the three angles inside a triangle is 180°. If is 24.9°, this means that the third angle in that triangle must be 25.1°. So:
(d) What is the critical angle for the diamond-air interface?
(e) What happens to the light at the diamond-air interface?
Because the angle of incidence (64.9°) is larger than the critical angle, the light is totally reflected internally.
(f) The light is reflected off the interface, obeying the law of reflection. It then strikes the diamond-water interface. What happens to it here?
Again, the place to start is by determining the angle of incidence
Because the angle of incidence is less than the critical angle, the beam will escape from the piece of diamond here. The angle of refraction can be found from Snell's law:
Because the angle of incidence (64.9°) is larger than the critical angle, the light is totally reflected internally.
(f) The light is reflected off the interface, obeying the law of reflection. It then strikes the diamond-water interface. What happens to it here?
Again, the place to start is by determining the angle of incidence
Because the angle of incidence is less than the critical angle, the beam will escape from the piece of diamond here. The angle of refraction can be found from Snell's law:
2 comments:
rohit beniwal this syd ma'am !! this is cool stuff..thgh we have'nt covered reflection in class yet ! thanks anyway !
the smallest smile is not lost
each wavelet on the ocean tossed
of life brings smile that flow and flow
each smiling lessens human woe
thanks for the assignment mam
sonam chhabra
xii-d
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