I passed green laser light from air to water but to my suprise the light doesn't bend towards normal even though I changed the angle of incidence.It goes undeflected ,Is it because of its higher frequency?enter image description here

Projecting Directly into water

  • $\begingroup$ It's supposed to refract: instructional-resources.physics.uiowa.edu/demos/…. Can you show how you set this up? $\endgroup$ Commented Mar 5 at 14:58
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Mar 5 at 15:17
  • $\begingroup$ It would be better to attach a picture of your result. $\endgroup$
    – Ankit
    Commented Mar 5 at 15:23
  • $\begingroup$ Does the shape of the medium containing water matters? $\endgroup$
    – Sanjay S
    Commented Mar 5 at 15:36
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    $\begingroup$ It looks to me like it does refract. You have it at almost 90 degrees so you only expect a little refraction and it looks to me like there is a little refraction. Also, this doesn’t look like water. $\endgroup$
    – Dale
    Commented Mar 5 at 18:11

2 Answers 2


It does refract,- never trust your "eyeballing", but use any possible measuring tool instead (even GIMP angle measure tool will do). The only problem is that we don't know surface normal vector at the incident point, so we have to guess it. If it is like here (rotated & cropped image; retouched a bit,- for in-out beams to match same entrance point; normal (probable) vector added; directions of beams added) :

enter image description here

Then according to a Snell's law, $$ \frac {\sin(16.5^\circ)}{\sin(11^\circ)} \approx 1.5 \approx n_{glass}$$

Water isn't floating in the air, but exists in some flask made of glass , right ? So laser beam must penetrate glass at first, not water, so you have to analyze beam refraction from air to the glass surface and only then from glass to the water. So laser beam actually makes double-refraction and that may cause the total effect you see.

In addition, we don't know full shape of glass, nor glass impurities at the contact site, nor defects (maybe there are cracks or unclean spots on the surface) which also tampers the refraction effect.

Lastly, I want to note, that science is not a simple thing. When performing experiment, you have to acquire total control of environment and conditions. Make sure that there are no undesired effects at the play, you have to cancel unneeded ones and so on until you are $100\%$ sure that you are testing required phenomena to a required accuracy. It's actually very hard,- especially when in the last century "all the simple science" was done and now to advance further we need even bigger laboratories/equipment which alas, means even bigger funding needed, which pity,- may not be granted for all science applications.

My this rant was to show that it's very easy to throw something out of the window and say "it doesn't work", but try to imagine how it was hard for the scientists to come up with this thing at the first place.

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    $\begingroup$ That was really helpful! Thanks for your time :) $\endgroup$
    – Sanjay S
    Commented Mar 6 at 1:53
  • $\begingroup$ I think so you made a mistake in calculating the angle of incidence. To point out exactly, you have drew the normal in a wrong way . I think you didn't consider the surface vector(which is 90° to the medium) $\endgroup$
    – Sanjay S
    Commented Mar 6 at 2:16
  • $\begingroup$ @SanjayS The basic point of my post is that there is a refraction, because one can see an angle between incident and outgoing beams (red lines). Now, surely exact refraction scheme is under consideration (this scheme was my interpretation), especially when picture is not very clean and it's hard to tell from it how beam refract. $\endgroup$ Commented Mar 6 at 6:54
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    $\begingroup$ Is the blue line supposed to be the normal? Honestly the laser beam looks a lot closer to normal than the blue line. This will reduce both the 16.5 and 11 degree measurements by a constant, completely changing the outcome of your calculation. As for the small angle between the beam inside and outside, that's also affected significantly by the curved surface that you're viewing the beam through. So I'm not sure you should be attempting to extract the index of refraction from this analysis. $\endgroup$
    – AXensen
    Commented Mar 6 at 19:03
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    $\begingroup$ we can't consider blue colour as normal right? Because it is not perpendicular to the surface, so the red one is normal . And the incident ray is 0° with normal . So, according to snell's law, n=1 , which shows no deviation or slight deviation (does that mean snell's law needs a constant to be added?). I made a mistake in not considering the surface while taking normal. $\endgroup$
    – Sanjay S
    Commented Mar 7 at 1:23

The difference in refractive index of water for red light versus green light is 1.332 versus 1.334 or so, so that is not the reason for not seeing any refraction.Optical properties of water

That means to understand what's happening you need to think through the details of your experiment more. Here are some questions

  1. How exactly are you determining how much refraction there is? Are you measuring it? Eyeballing it?
  2. Are you allowing the beam to pass back out of the water before checking the amount of deflection? If so, you won't see any deflection at all.
  3. Are you trying a full range of angles of incidence? The deflection of the beam should be less than 10 degrees for all angles less than 40 degrees. To really see obvious refraction you need larger angles of incidence.

EDIT: It sounds to me like you have not really made a plan for how you will measure whether there is refraction or not and so you are doing things that make it difficult to see the refraction. For example, when you look at the water the only way to tell if there is refraction is if the beam hits the bottom of the container in a different place than if it traveled in a straight line. That's very hard to eyeball, especially if you are dealing with small amounts of deflection. And you probably are dealing with small amounts of deflection because you are shining the beam into the top of a narrow container.

To really test the refraction in these situations you need to come up with a way of measuring the refraction that is more precise than "eyeballing."

Images of problems with the current set up

  • $\begingroup$ 1.eye balling. 2.checking the water in the sense? 3. Isn't it goes undeflected only in the angle of 90° $\endgroup$
    – Sanjay S
    Commented Mar 5 at 15:13
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    $\begingroup$ 2. Sorry, "checking the refraction." Edited. If a beam goes into water and then back out into air the intial deflection gets undone and it comes out parallel. $\endgroup$ Commented Mar 5 at 15:38
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    $\begingroup$ 3. Yes, the beam is only technically undeflected when it hits the water head on, but if you are eyeballing the amount of deflection can you actually see 5 degrees of deflection? You may think there is no deflection when there actually is. If you really want to see the refraction you need to measure it $\endgroup$ Commented Mar 5 at 15:40
  • $\begingroup$ Oh, thanks for your reply! I have an another question. I observed refraction when I pass light from outside the glass container that has water in it(I observed laterally). But when I pass it directly into water I can't find deflection,is there any specific way to look for refraction?(In case 2 I was looking from the top) $\endgroup$
    – Sanjay S
    Commented Mar 5 at 15:47
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    $\begingroup$ Without a more detailed description of what you are doing and what you are looking at it is difficult to give advice. $\endgroup$ Commented Mar 5 at 15:51

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