Electromagnetic radiation wavelength

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SUMMARY

The discussion focuses on calculating the wavelength of electromagnetic radiation with a frequency of 5.00x1014 Hz. The wavelength in a vacuum is determined to be 600 nm (6.00x10-7 m), while in water, with an index of refraction of 1.33, the wavelength is calculated to be 451 nm. This radiation is confirmed to be visible, appearing orange at 600 nm. Additionally, the index of refraction of a medium where the speed of light is 2.54x108 m/s is discussed, emphasizing that the frequency remains constant while the wavelength changes across different media.

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An Electromagnetic radiation has a frequency of 5.00x10^14 hx.
a) Calc its wavelength in a vacuum in metres and nanometres
b)Calc its wavelength in water
c)is this radiation visible? if so wat colour?
d) What is the index of refraction of a medium in which the speed of this radiation is 2.54x10^8m/s?
e)where would u encounter this radiation in your daily life?

My attempt...

a)c=f(wavelength) wavelength = 3x10^8/5x10^8
=600nm and 6.00x10^-7 m

b)n=1.33 in water ?=6.00x10^-7/1.33
wavelength =600nm =451nm
wavelength in median ??

c)yes this radiation is visible and according to the visible spectrum, 590-610nm is orange so its at 600nm so it is orange.

d)i cannot do this question, can anyone help me??

e)same with this one
 
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d)c=f(wavelength)
2.54x10^8/5x10^14=wavelenngth
=5.08x10^-7m or 508 nm but that is now totally different from my other wavelength in a) this question is confusing...please help
 
The index of refraction of a medium is the ratio of the speed of light in vacuum (air) to its speed in the medium.

The frequency stays the same in both media, it is the wavelength that changes. Your ratio is the wrong way round.

Traffic light?
 
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