A monochromatic light of frequency \( 5 \times 10^{14} \) Hz travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :
Show Hint
- 300 nm
- 600 nm
- 400 nm
- 500 nm
The Correct Option is A
Approach Solution - 1
The frequency of light remains unchanged when it passes from one medium to another.
Given frequency \( f = 5 \times 10^{14} \) Hz.
The speed of light in air (approximately vacuum) is \( c = 3 \times 10^8 \) m/s.
The wavelength of light in air (\( \lambda_{air} \)) is given by: \[ \lambda_{air} = \frac{c}{f} = \frac{3 \times 10^8 \, \text{m/s}}{5 \times 10^{14} \, \text{Hz}} = 0.6 \times 10^{-6} \, \text{m} = 600 \times 10^{-9} \, \text{m} = 600 \, \text{nm} \] The refractive index of the medium is given as \( \mu = 2 \). The wavelength of light in the medium (\( \lambda_{medium} \)) is related to the wavelength in vacuum (or air) by the refractive index: \[ \lambda_{medium} = \frac{\lambda_{air}}{\mu} \] Substituting the values: \[ \lambda_{medium} = \frac{600 \, \text{nm}}{2} = 300 \, \text{nm} \] The wavelength of the refracted light in the medium is 300 nm.
Approach Solution -2
To solve the problem of finding the wavelength of refracted light when monochromatic light travels from air into a medium of refractive index 2, we can use the formula that relates wavelength, frequency, and the speed of light.
First, recall the relationship between speed, frequency, and wavelength:
\(v = \nu \lambda\)
Where:
- \(v\) is the speed of light in the medium.
- \(\nu\) is the frequency of the light.
- \(\lambda\) is the wavelength of the light.
The speed of light in a medium is given by:
\(v = \frac{c}{n}\)
Where:
- \(c \approx 3 \times 10^8 \, \text{m/s}\) is the speed of light in vacuum.
- \(n\) is the refractive index of the medium.
Given:
- Frequency \(\nu = 5 \times 10^{14} \, \text{Hz}\)
- Refractive index \(n = 2\)
Substitute \(v\) in the wavelength formula:
\(\frac{c}{n} = \nu \lambda\_m\)
Rearrange to solve for \(\lambda\_m\) (wavelength in the medium):
\(\lambda\_m = \frac{c}{n \nu}\)
Substitute the known values:
\(\lambda\_m = \frac{3 \times 10^8}{2 \times 5 \times 10^{14}} = 3 \times 10^{-7} \, \text{m}\)
Convert meters to nanometers (1 m = 109 nm):
\(\lambda\_m = 3 \times 10^{-7} \times 10^9 = 300 \, \text{nm}\)
Thus, the wavelength of the refracted light is 300 nm, which is the correct answer.
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