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Question: When the orange light passes from air (\[n=1\]) into glass (\(n=1.5\)), what is its new wavelength? ...

When the orange light passes from air (n=1n=1) into glass (n=1.5n=1.5), what is its new wavelength?
A.4×10−7mA.4\times {{10}^{-7}}m
B.B. 4×10−6m4\times {{10}^{-6}}m
C.2.5×10−7mC.2.5\times {{10}^{-7}}m
D.6×10−7mD.6\times {{10}^{-7}}m
E.9×10−7mE.9\times {{10}^{-7}}m

Explanation

Solution

We will use the concept of absolute refractive index and formula of wavelength in terms of frequency and speed of light. We also know that when a light ray travels from one medium to another then the wavelength and speed of light change but frequency remains the same.
Formula Used: We will use the following formula to solve the given problem:-
cn=νλ\dfrac{c}{n}=\nu \lambda .

Complete step by step answer:
We know that the absolute refractive index of a medium is defined as the ratio of speed of light in vacuum to the speed of light in the medium. If cc is speed of light in vacuum, vv is speed of light in the medium and nn is the absolute refractive index of the medium, then mathematically
n=cvn=\dfrac{c}{v}………………. (i)(i)
From (i)(i) we can write
v=cnv=\dfrac{c}{n}……………… (ii)(ii)
But for velocity of light in a medium can also be represented with the help of wavelength (λ)(\lambda ) and frequency (ν)(\nu ) by following equation:-
v=ν×λv=\nu \times \lambda …………………. (iii)(iii)
For air we have velocity of light, vvequal to cc and wavelength of orange light, λa=600nm=600×10−9m{{\lambda }_{a}}=600nm=600\times {{10}^{-9}}m.
From equation (ii)(ii) and (iii)(iii) for air we get
cna=ν×λa\dfrac{c}{{{n}_{a}}}=\nu \times {{\lambda }_{a}}……………. (iv)(iv) where the refractive index of air is denoted as na=1{{n}_{a}}=1.
Putting respective values in (iv)\left( iv \right), we get
c1=ν×600×10−9\dfrac{c}{1}=\nu \times 600\times {{10}^{-9}}
⇒ν=c600×10−9\Rightarrow \nu =\dfrac{c}{600\times {{10}^{-9}}}………………… (v)(v)
For glass we have wavelength,λg{{\lambda }_{g}} which is to be calculated. Refractive index, ng=1.5{{n}_{g}}=1.5 and frequency of the orange light will remain the same as it does not change in different media. We have the following equation on the basis of equation (ii)(ii) and (iii)\left( iii \right):-
cng=ν×λg\dfrac{c}{{{n}_{g}}}=\nu \times {{\lambda }_{g}}…………………….. (vi)(vi)
Putting respective values in (vi)(vi), we get
c1.5=ν×λg\dfrac{c}{1.5}=\nu \times {{\lambda }_{g}}
⇒ν=c1.5×λg\Rightarrow \nu =\dfrac{c}{1.5\times {{\lambda }_{g}}}……………….. (vii)(vii)
We know that the frequency of light is same in both the cases and therefore from equations (v)(v) and (vii)(vii), we get
c600×10−9=c1.5×λg\dfrac{c}{600\times {{10}^{-9}}}=\dfrac{c}{1.5\times {{\lambda }_{g}}}
Solving further we get
600×10−9=1.5×λg600\times {{10}^{-9}}=1.5\times {{\lambda }_{g}}
⇒λg=600×10−91.5\Rightarrow {{\lambda }_{g}}=\dfrac{600\times {{10}^{-9}}}{1.5}
⇒λg=400×10−9m\Rightarrow {{\lambda }_{g}}=400\times {{10}^{-9}}m
⇒λg=4×10−7m\Rightarrow {{\lambda }_{g}}=4\times {{10}^{-7}}m

Hence, the correct option is (A)(A).

Note:
We should keep in mind that the frequency of light does not change with change in medium. We should be clear in the concept of absolute refractive index and should not consider it as a relative refractive index. Correct formula should be used and colour of the light should also be taken care of.