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Snell’s Law AQA A-Level Physics

I have got a bit stuck on this question so I wondered if somebody could explain it to me! I have completed part A correctly but am struggling to understand how I would calculate the refractive index for a water wave!Water waves of frequency 4.0 Hz travelling at a speed of 0.16 m/s travel across a boundary from deep to shallow water where the speed is 0.12 m/s. A) Calculate the wavelength of these wavesi) in the deep water (0.04M)i) in the shallow water (0.03M)B) The incident wavefronts cross the boundary at an angle of 25° to the boundary. Calculate the angle of the refracted wavefronts to the boundary.Thank You
What have you tried so far? If you know Snell's law and you know how refractive index and speed (or wavelength, as both same frequency) are related, then it should just be a simple rearrangement. Hint: how do you determine the relative refractive index at the boundary?
Original post by username6497536
I have got a bit stuck on this question so I wondered if somebody could explain it to me! I have completed part A correctly but am struggling to understand how I would calculate the refractive index for a water wave!Water waves of frequency 4.0 Hz travelling at a speed of 0.16 m/s travel across a boundary from deep to shallow water where the speed is 0.12 m/s. A) Calculate the wavelength of these wavesi) in the deep water (0.04M)i) in the shallow water (0.03M)B) The incident wavefronts cross the boundary at an angle of 25° to the boundary. Calculate the angle of the refracted wavefronts to the boundary.Thank You
im also stuck. i managed to work it out and get the correct annswer however the question states that the angle given is 25 degrees to the boundary. I assumed this meant this is not the incident angle and that i had to work out the incident angle by subtracting 90 degrees from the 25 degrees. When i worked the answer out by finding the incident angle it is incorrect. Does this mean i completly interpreted the question wrong and the angle given was in fact the incident angle?
I've made a diagram if that helps - DW for deep water and SW for shallow water. Since its 25 degrees to the boundary you need to 90-25 to work out angle from normal and then do snells law.
Original post by usename1234
im also stuck. i managed to work it out and get the correct annswer however the question states that the angle given is 25 degrees to the boundary. I assumed this meant this is not the incident angle and that i had to work out the incident angle by subtracting 90 degrees from the 25 degrees. When i worked the answer out by finding the incident angle it is incorrect. Does this mean i completly interpreted the question wrong and the angle given was in fact the incident angle?

TSR_Snell’s Law AQA A-Level Physics.jpg
The drawing (as shown above) created by mosaurlodon in post # 4 is wrong!

What is the correct drawing?
Follows this link. See Figure 6.
https://www.coastalwiki.org/wiki/Shallow-water_wave_theory#Refraction

Use Figure 6 from the link, you can work out using the info (“The incident wavefronts cross the boundary at an angle of 25° to the boundary”) with some geometry arguments to show that 25° is indeed the incident angle in the deep water.
(edited 1 month ago)
Water waves of frequency 4.0 Hz travelling at a speed of 0.16 m/s travel across a boundary from deep to shallow water where the speed is 0.12 m/s.
A) Calculate the wavelength of these waves
i) in the deep water (0.04M)
ii) in the shallow water (0.03M)
B) The incident wavefronts cross the boundary at an angle of 25° to the boundary. Calculate the angle of the refracted wavefronts to the boundary.




This post is for people who are reading this thread who have similar problems in future or now.

The diagram that describes the problem is shown below.



The info “The incident wavefronts cross the boundary at an angle of 25° to the boundary” tells us that angle α0 = 25° in the given diagram, so the angle of the refracted wavefronts to the boundary is α.

We can compute α using Snell’s law (equation (12)) given in the following link without the concept of refractive index. As far as I know, we do not talk about the refractive index for water waves. I would not recommend that anyone use such a concept.
https://www.coastalwiki.org/wiki/Shallow-water_wave_theory#Refraction

Spoiler



Just a side note: We can show α0 and α are the angles of incidence and refraction respectively by using some geometry arguments or right-angled triangles.
(edited 1 month ago)
Original post by Eimmanuel
This post is for people who are reading this thread who have similar problems in future or now.

The diagram that describes the problem is shown below.



The info “The incident wavefronts cross the boundary at an angle of 25° to the boundary” tells us that angle α0 = 25° in the given diagram, so the angle of the refracted wavefronts to the boundary is α.

We can compute α using Snell’s law (equation (12)) given in the following link without the concept of refractive index. As far as I know, we do not talk about the refractive index for water waves. I would recommend that anyone use such a concept.
https://www.coastalwiki.org/wiki/Shallow-water_wave_theory#Refraction

Spoiler



Just a side note: We can show α0 and α are the angles of incidence and refraction respectively by using some geometry arguments or right-angled triangles.
Thanks, i understand now. The diagram reminds me of an object on a slope of angle θ so i was able to get it instantly yayyyy! :smile:))))
Oh so I had the complete wrong method - my bad bro

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