Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The wavelength of blue light in air is \[4500{}^\circ A\] . If the refractive index for blue light is \[1.55\] in glass, the frequency of the refracted light in glass and its wavelength are respectively \[(c=3\times {{10}^{8}}m/s)\] :
(A) \[6.7\times {{10}^{14}}Hz,3000A{}^\circ \]
(B) \[7.5\times {{10}^{14}}Hz,7000A{}^\circ \]
(C) \[4.5\times {{10}^{14}}Hz,3500A{}^\circ \]
(D) \[5.4\times {{10}^{14}}Hz,7000A{}^\circ \]

seo-qna
Last updated date: 24th Jun 2024
Total views: 405k
Views today: 4.05k
Answer
VerifiedVerified
405k+ views
Hint: Refractive index for a medium is the ratio of the speed of light in air to speed of light in that medium. Using this relation, we can find the speed of light in the given medium. Once we have the speed of light, we can make use of the relation between the speed of light, the frequency of light and the wavelength of light to find the other result. Also, the frequency of light does not change during refraction.

Formula Used:
\[\mu =\dfrac{c}{v}\]
\[c=\lambda \times \eta \]

Complete step by step answer:
As discussed in the hint, the refractive index of a medium is equal to the ratio of the speed of light in air to the speed of light in the given medium.
Mathematically, \[\mu =\dfrac{c}{v}\] where \[\mu \] is the refractive index of blue light in light, \[c\] is the speed of blue light in air and \[v\] is the speed of blue light in the glass.
Substituting the values in the above equation, we get
\[\begin{align}
  & 1.55=\dfrac{3\times {{10}^{8}}m/s}{v} \\
 & \Rightarrow v=\dfrac{3\times {{10}^{8}}}{1.55}m/s \\
 & \Rightarrow v=1.94\times {{10}^{8}}m/s \\
\end{align}\]
Now, coming back to the relation between the velocity, frequency and wavelength of light, we know that the speed of light is equal to the product of the wavelength and the frequency.
Mathematically, \[c=\lambda \times \eta \] where \[c\] is the speed of blue light in air, \[\lambda \] is the wavelength of the light and \[\eta \] is the frequency of the light.
Substituting the values, we get
\[\begin{align}
  & 3\times {{10}^{8}}m/s=4500{}^\circ A\times \eta \\
 & \Rightarrow \eta =\dfrac{3\times {{10}^{8}}m/s}{4500{}^\circ A} \\
 & \Rightarrow \eta =\dfrac{3\times {{10}^{8}}m/s}{4500\times {{10}^{-10}}m}(\because 1A{}^\circ ={{10}^{-10}}m) \\
 & \Rightarrow \eta =\dfrac{3}{45}\times {{10}^{16}}Hz=0.067\times {{10}^{16}}Hz \\
 & \Rightarrow \eta =6.7\times {{10}^{14}}Hz \\
\end{align}\]
Since we know that frequency of the light does not change on refraction and we know the velocity of light in the glass, we can easily find the wavelength of light in the glass.
Substituting the values in the relation \[v=\lambda \times \eta \] , we get
\[\begin{align}
  & 1.94\times {{10}^{8}}m/s=\lambda \times 6.7\times {{10}^{14}}Hz \\
 & \Rightarrow \lambda =\dfrac{1.94\times {{10}^{8}}m/s}{6.7\times {{10}^{14}}Hz} \\
 & \Rightarrow \lambda =0.2895\times {{10}^{-6}}m \\
 & \Rightarrow \lambda =2895A{}^\circ (\because 1A{}^\circ ={{10}^{-10}}m) \\
 & \Rightarrow \lambda \simeq 3000A{}^\circ \\
\end{align}\]

Hence the correct option is (A).

Note: If we use a little trick, we can avoid a lot of effort in our calculation in this question. We can use the process of option elimination. We can easily calculate the frequency of light in air using the relation between the speed, wavelength and frequency, that is \[v=\lambda \times \eta \] . Now, if we know the simple fact that the frequency won’t change on refraction, we can skip the remaining calculation and check the options straight away and eliminate all the wrong options.