Question

Refractive Index of a certain glass is 1.5 for light whose wavelength in vacuum is 6000 angstrom the wavelength of this light when it passes through glass is

Answer 1

Wavelength of light in air = 6000A°= 6000A°×10^-8 cm/A° = 6× 10^-5 cm

The velocity of light in air= 3×10^10 cm/s

Frequency v of light in air (velocity of light in air in cm/s) ÷ wave length of light in cm= ( 3×10^10 cm/s)/6×10^-5 cm= 5×10^14 Hz.

On entering the medium of refractive index n= 1.5 from air, the velocity of light decreases to c'= c/n= 3×10^10 cm per sec./ 1.5 = 2×10^10 cm/s.

The frequency of light does not change on refraction. So frequency of light in the medium= 5×10^14 Hz.

To keep the frequency unchanged, the wavelength of light in the medium also decreases by the same factor as the decrease in the velocity of light. So altered wavelength of light in the medium= 6000A°/1.5 = 4000A° .

So frequency of light in the medium= 5×10^14 Hz.

Wavelength of light in the medium= 4000A° .

Answer 2

Given, the refractive index $\mu = 1.5$

Given wavelength in vacuum $= 6000\AA$

Now, we know that $\mu = \frac{c}{v}$,

where c is the speed of light and v is the speed of light in that medium.

$\therefore \mu=\frac{ \lambda_{vacuum} \nu_{vacuum}}{\lambda_{medium} \nu_{medium}}$

As frequency $\nu$ remains constant irrespective of medium,

$\mu = \frac{\lambda_{vacuum}}{\lambda_{glass}}=\frac{ 6000}{\lambda_{glass}}=1.5\\\therefore \lambda_{glass}= \frac{6000}{1.5}= 4000\AA$

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