Light of wavelength 590nm in air enters into water. Find (I) speed (ii) frequency (iii) wavelength of light in water. Assume that frequency of light remain same in both the media and refractive index of water is 1.33.
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1)We know that the refractive index= velocity of light in vacuum÷ Velocityof light in medium
Therefore, 1.33= 3*10^8 ÷ velocity of light in water
Velocity of light in water=3*10^8/1.33
=2.25*10^8 m/sec
2)The frequency of light remains unchanged even if the medium is changed
Velocity = frequency * wavelength
3*10^8 = frequency * 590*10^-9
Frequency = 5.08*10^14 Hz
3) wavelength of light in air ÷ wavelength of light in water = refractive index
Wavelength of light in water = 590*10^-9/1.33=4.436nm
Therefore, 1.33= 3*10^8 ÷ velocity of light in water
Velocity of light in water=3*10^8/1.33
=2.25*10^8 m/sec
2)The frequency of light remains unchanged even if the medium is changed
Velocity = frequency * wavelength
3*10^8 = frequency * 590*10^-9
Frequency = 5.08*10^14 Hz
3) wavelength of light in air ÷ wavelength of light in water = refractive index
Wavelength of light in water = 590*10^-9/1.33=4.436nm
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