GCSE Physics: Refraction through a semi-circular block practical

  Рет қаралды 22,401

Burrows Physics

Burrows Physics

Күн бұрын

A demo showing how light refracts through a semi-circular glass block and how Snell's law is used to model it.

Пікірлер: 20
@dngosama
@dngosama 3 жыл бұрын
Thank you I understood 🥰
@LoisBolling
@LoisBolling Жыл бұрын
Hey, thanks for the demonstration! I'm a huge Pink Floyd fan and I wanted to recreate the refraction of light with a glass prism. I was wondering what kind of light source you were using to get that thin beam. As far as I know, to recreate the full color spectrum that sunlight offers, you need a light source capable of producing those colors. Is it an incandescent light bulb that you're using, or something else? Thanks a lot!
@burrowsphysics
@burrowsphysics Жыл бұрын
I'll reply on your other comment!
@qwertyuui0762
@qwertyuui0762 Жыл бұрын
thank u for this:)
@dia8203
@dia8203 2 жыл бұрын
Would it make any difference if the light passes through the other side of the glass? By that i mean the straight side instead of the curved side.
@burrowsphysics
@burrowsphysics 2 жыл бұрын
The light paths you see are completely reversible. So if you place the ray box at the other end of the light ray, it would trace an identical path in the opposite direction
@ajithm9459
@ajithm9459 3 жыл бұрын
How does the focal length of the semicircular lens vary with wavelength?can any one answer!!pls
@burrowsphysics
@burrowsphysics 3 жыл бұрын
Violet light has a larger direction change when it is refracted through a boundary which to me would suggest it would have a shorter focal length. Interesting question though I'm not certain about that
@malihahijab6466
@malihahijab6466 2 жыл бұрын
how the narrow beam of light is accurate to find the path of ray through the block?
@burrowsphysics
@burrowsphysics 2 жыл бұрын
There isn't a specific answer to that. Generally speaking the narrower the better for accuracy of measurements but then again as the beam gets narrower it is harder to see
@notson_of_pears6024
@notson_of_pears6024 2 жыл бұрын
Not sure if your going to reply in a enough time since my exam is in 2 days, however, if you're able to reach back, I'm wondering why the light is not bent towards the normal. Is this because of the different shape of the Perspex?
@burrowsphysics
@burrowsphysics 2 жыл бұрын
I'm going to need some context to answer your question! In what circumstance specifically are you referring to?
@ScubaShark--8964
@ScubaShark--8964 Жыл бұрын
@@burrowsphysics Shouldn't 0.6/0.93=0.64516129 (if 2dp) be 0.65 instead of 0.64?
@cynical5150
@cynical5150 Жыл бұрын
@@burrowsphysics i think hes talking about the light, when it goes from a less dense medium to a denser medium it will bend towards the normal (air to glass semicircle thingy). But during the experiment the light only refracts at the the centre. Why? I also wonder. The wavelength of light is changing therefore the direction of light should change as well. Thats just what im thinking
@burrowsphysics
@burrowsphysics Жыл бұрын
At the other boundary the light is travelling along the normal so it is unrefracted even though its speed changes. (From your answer it sounds like you think the change in wavelength causes refraction but actual it is the change in speed that causes it, the change in wavelength is just a bi-product of the speed change because frequency stays the same)
@tanchunrui2550
@tanchunrui2550 Жыл бұрын
I wonder if the refractive index can be less than 1?
@burrowsphysics
@burrowsphysics Жыл бұрын
No it can't because that would mean light was travelling faster than the speed of light in a vacuum which is impossible
@IbrahimKhan-qz9df
@IbrahimKhan-qz9df Жыл бұрын
I have been here :)
@butterfly-fx2tl
@butterfly-fx2tl 3 жыл бұрын
Wait, I didn't understand why the light ray didn't get refracted as it passed through the curved side of the semi-circular glass block. Can you explain it?
@burrowsphysics
@burrowsphysics 3 жыл бұрын
As the ray was travelling along the normal to the boundary, it had an angle of incidence of zero and so was unrefracted.
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