
The refractive index of a prism is measured using three lines of a mercury vapor lamp. If \[\mu 1,\mu 2\]and \[\mu 3~\] are the measured refractive indices for these green, blue and yellow lines respectively, then
A. \[\mu 2>\mu 3~>\mu 1\]
B. \[\mu 2>\mu 1~>\mu 3\]
C. \[\mu 3>\mu 2>\mu 1\]
D. \[\mu 1>\mu 2~>\mu 3\]
Answer
591.3k+ views
Hint: Waves tend to change the direction while passing from one medium to another or when there is gradual change in a medium. This process is called refraction. So when mercury vapour lamp lines pass from air to prism, a bend in light occurs as the rays move from rarer to denser medium.
Formula used: Cauchy’s equation for refractive index of a material
\[\mu (\lambda )=A+\dfrac{B}{{{\lambda }^{2}}}+\dfrac{C}{{{\lambda }^{4}}}.......\]
Where A,B,C, are constants . Since wavelength has very small values, the higher powers are neglected and thus,
\[\mu (\lambda )=A+\dfrac{B}{{{\lambda }^{2}}}\]
Complete step by step answer:
Refractive index is the value that is generated from the ratio between the speed of light in vacuum to that in a second medium with greater density. Its variable is symbolized by the letter n or n' .
Refractive index n is obtained by
$n=\dfrac{c}{v}$
When we apply Cauchy’s equation for refractive index of a material, that is,
\[\mu (\lambda )=A+\dfrac{B}{{{\lambda }^{2}}}\] where $B>0$
We observe that as the wavelength increases, the refractive index decreases.
We know that in the visible spectrum, the order of wavelength is given by VIBGYOR and it follows ascending order.
The lines mentioned here are green, blue and yellow.
$\Rightarrow \lambda B<{{\lambda }_{G}}<{{\lambda }_{Y}}$ i.e. ${{\lambda }_{2}}<{{\lambda }_{1}}<{{\lambda }_{3}}$
Thus, we can say that, \[\mu 2>\mu 1~>\mu 3\]
So, the correct answer is Option B .
Note:
Refractive index has abundant application purposes. It is widely utilised for identification of a particular substance, to check its purity and or to analyse its concentration. Usually it is used to measure the concentration of a solute in aqueous solutions.
Formula used: Cauchy’s equation for refractive index of a material
\[\mu (\lambda )=A+\dfrac{B}{{{\lambda }^{2}}}+\dfrac{C}{{{\lambda }^{4}}}.......\]
Where A,B,C, are constants . Since wavelength has very small values, the higher powers are neglected and thus,
\[\mu (\lambda )=A+\dfrac{B}{{{\lambda }^{2}}}\]
Complete step by step answer:
Refractive index is the value that is generated from the ratio between the speed of light in vacuum to that in a second medium with greater density. Its variable is symbolized by the letter n or n' .
Refractive index n is obtained by
$n=\dfrac{c}{v}$
When we apply Cauchy’s equation for refractive index of a material, that is,
\[\mu (\lambda )=A+\dfrac{B}{{{\lambda }^{2}}}\] where $B>0$
We observe that as the wavelength increases, the refractive index decreases.
We know that in the visible spectrum, the order of wavelength is given by VIBGYOR and it follows ascending order.
The lines mentioned here are green, blue and yellow.
$\Rightarrow \lambda B<{{\lambda }_{G}}<{{\lambda }_{Y}}$ i.e. ${{\lambda }_{2}}<{{\lambda }_{1}}<{{\lambda }_{3}}$
Thus, we can say that, \[\mu 2>\mu 1~>\mu 3\]
So, the correct answer is Option B .
Note:
Refractive index has abundant application purposes. It is widely utilised for identification of a particular substance, to check its purity and or to analyse its concentration. Usually it is used to measure the concentration of a solute in aqueous solutions.
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