Parallel beam containing light of $\lambda = 400$nm and 500 nm is incident on a prism as shown in figure. The refractive index $\mu$ of the prism is given by the relation $\mu(\lambda) = 1.20 + \frac{0.8 imes 10^{- 14}}{\lambda^{2}}$ Which of the following statement is correct? [image:9_image20.emf]

Question

Parallel beam containing light of $\lambda = 400$nm and 500 nm is incident on a prism as shown in figure. The refractive index $\mu$ of the prism is given by the relation $\mu(\lambda) = 1.20 + \frac{0.8 	imes 10^{- 14}}{\lambda^{2}}$ Which of the following statement is correct? [image:9_image20.emf]

Dr. Abubacker · Accepted Answer

(a) $\mu_{1} = 1.20 + \frac{0.8 	imes 10^{- 14}}{\left( 400 	imes 10^{- 9} ight)^{2}}$
Or $\mu_{1} = 1.20 + \frac{0.8 	imes 10^{- 14}}{400 	imes 400 	imes 10^{- 18}}$
Or $\mu_{1} = 1.20 + \frac{0.8}{16}$
Or $\mu_{1} = 1.20 + 0.05$
Or $\ \mu_{1} = 1.25$
Or $\ \ \sin{i_{c} = \frac{1}{1.25} = 0.8}$
Or $i_{c} = 53.13{^\circ}$
Again,${\ \ \mu}_{2} = 1.20 + \frac{0.8 	imes 10^{- 14}}{\left( 500 	imes 10^{- 9} ight)^{2}}$
[image:9_image85.jpeg]
Or $\ \mu_{2} = 1.20 + \frac{0.8}{25}$ or $\mu_{2} = 1.20 + 0.32$
Or ${\ \mu}_{2} = 1.232$
Or $\sin{i_{c} = \frac{1}{1.232} = 0.81}$
Or $\ i_{c} = \sin^{- 1}{0.81}$
$= 54.26{^\circ}$
Now,${\ \ sin}{	heta = 0.8\ or\ 	heta = 53.13{^\circ}}$
This angle is clearly greater than critical angle corresponding to wavelength 400 nm. So, light of 400 nm wavelength under goes total internal reflection

Q. Parallel beam containing light of $\lambda = 400$nm and 500 nm is incident on a prism as shown in figure. The refractive index $\mu$ of the prism is given by the relation $\mu(\lambda) = 1.20 + \frac{0.8 \times 10^{- 14}}{\lambda^{2}}$ Which of the following statement is correct?

Explanation

Reviewed by Dr. Abubacker

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