Hi Farhaini,
SPM Physics 2005 Q25 P1 Light & Optics Question
25 The diagram shows a light ray directed into a glass block. Which is the angle of refraction? Gambar rajah menunjukkan satu sinar cahaya ditujukan kepada satu bongkah kaca. Sudut yang manakah adalah sudut biasan?
SPM Physics 2005 Q25 P1 Light & Optics Solution
The lower boundary of the glass block and upper boundary of the glass block should be viewed as 2 separate things. When the light ray go through the 2 boundaries, it has an incident and refracted angles for each boundaries separately.
C is incident angle, instead of angle of refraction. Because there is where the light ray enter the upper boundary, and get refracted.
I think incident and refracted angle should depends on the direction of light ray.
The point where light ray enter the boundary is always incident angle
The point where light ray exit the boundary is always refracted angle.
To find refractive index of glass with Snell’s law, it must always be angle in the air at the top, angle in the glass at the bottom.
When sometimes the light ray come out of the glass, I will imagine the light ray move in the opposite direction like above diagram. The way the light ray being refracted (angles etc.) will be the same, even it is in the opposite direction. Then Snell’s law can be applied with sin(i) / sin(r)
Ooh. Now I understand. Thank you very much for helping me a lott. I really appreciate it. You have my deepest thanks.
You are very welcome