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Question: Green light of wavelength \(5460A{}^\circ \) is incident on an air-glass interface. If the refractiv...

Green light of wavelength 5460A5460A{}^\circ is incident on an air-glass interface. If the refractive index of glass is 1.51.5, the wavelength of light in glass would be (given-velocity of light in air c=3×108ms1c=3\times {{10}^{8}}m{{s}^{-1}})
A)3460A B)5460A C)4861A \begin{aligned} & A)3460A{}^\circ \\\ & B)5460A{}^\circ \\\ & C)4861A{}^\circ \\\ \end{aligned}
D)D)None of the above

Explanation

Solution

Frequency of light is independent of the medium of propagation. Firstly, the frequency of green light is determined using the given wavelength of green light in air. Using the refractive index of glass, velocity of green light in glass is calculated. Finally, using the obtained frequency and velocity of green light in glass, the wavelength of green light inside the glass is calculated.
Formula used:
1)f=cλ1)f=\dfrac{c}{\lambda }
2)v=cn2)v=\dfrac{c}{n}
3)v=fλ3)v=f\lambda

Complete answer:
Frequency of light does not depend on the medium through which the light is travelling. It is dependent only on the source of light. Mathematically, frequency of light is given by
f=cλf=\dfrac{c}{\lambda }
where
ff is the frequency of light
cc is the speed of light in air
λ\lambda is the wavelength of light in air
Let this be equation 1.
Coming to our question, we are given that green light of wavelength 5460A5460A{}^\circ is incident on an air-glass interface. If the refractive index of glass is 1.51.5 and velocity of light in air is 3×108ms13\times {{10}^{8}}m{{s}^{-1}}, we are required to determine the wavelength of green light in glass.
Let us call the wavelength of green light in air λ1{{\lambda }_{1}}. Clearly,
λ1=5460A=5460×1010m{{\lambda }_{1}}=5460A{}^\circ =5460\times {{10}^{-10}}m
If we call the frequency of green light ff, using equation 1, we have
f=cλ1=3×108ms15460×1010m=5.49×1014Hz=5.49×1014s1f=\dfrac{c}{{{\lambda }_{1}}}=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{5460\times {{10}^{-10}}m}=5.49\times {{10}^{14}}Hz=5.49\times {{10}^{14}}{{s}^{-1}}
where
ff is the frequency of green light
cc is the speed of light in air (=3×108ms1=3\times {{10}^{8}}m{{s}^{-1}}, as given in the question)
λ1{{\lambda }_{1}} is the wavelength of green light in air
Let this be equation 2.
Now, we know that velocity of light in a particular material or medium is given by
v=cnv=\dfrac{c}{n}
where
nn is the refractive index of a material or a medium
cc is the speed of light in air
vv is the velocity of light in that material or that medium
Let this be equation 3.
Using equation 3, let us determine the velocity of green light in glass. Clearly,
v=cn=3×108ms11.5=2×108ms1v=\dfrac{c}{n}=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{1.5}=2\times {{10}^{8}}m{{s}^{-1}}
where
n=1.5n=1.5, is the refractive index of glass (as given in the question)
c=3×108ms1c=3\times {{10}^{8}}m{{s}^{-1}}, is the speed of light in air (as given in the question)
vv is the velocity of green light in glass
Let this be equation 4.
Velocity of light in a material is also equal to the product of frequency of light propagating through the material and the wavelength of light propagating through the material. Clearly,
v=fλv=f\lambda
where
vv is the velocity of light propagating through a particular material
ff is the frequency of light
λ\lambda is the wavelength of light in that particular material
Let this be equation 5.
Using equation 5, let us calculate the wavelength of green light in glass. If wavelength of green light in glass is denoted as λ2{{\lambda }_{2}}, then, using equation 5, λ2{{\lambda }_{2}} is given by
v=fλ22×108ms1=5.49×1014s1×λ2λ2=2×108ms15.49×1014s1=3642Av=f{{\lambda }_{2}}\Rightarrow 2\times {{10}^{8}}m{{s}^{-1}}=5.49\times {{10}^{14}}{{s}^{-1}}\times {{\lambda }_{2}}\Rightarrow {{\lambda }_{2}}=\dfrac{2\times {{10}^{8}}m{{s}^{-1}}}{5.49\times {{10}^{14}}{{s}^{-1}}}=3642A{}^\circ
Therefore, the wavelength of green light in glass is equal to 3642A3642A{}^\circ .

Since this answer is not there in the first three options provided, option DD is the correct answer.

Note:
Students need to be thorough with conversion formulas. Conversion formulas used in the solution given above are:
1A=1010m 1Hz=1s1 \begin{aligned} & 1A{}^\circ ={{10}^{-10}}m \\\ & 1Hz=1{{s}^{-1}} \\\ \end{aligned}
Students should also try to remember that frequency of light is independent of the medium of propagation whereas wavelength of light is dependent on the medium of propagation.