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, $\mu(\lambda)=1.20+\frac{0.8 \times 10^{-14}}{\lambda^{2}}$ Which of the following statement is correct?

• A. Light of $\lambda=400\, nm$ undergoes
• B. Light of $\lambda=500\, nm$ undergoes total internal reflection
• C. Neither of the two wavelengths undergoes total intemal reflection
• D. Both wavelengths undergoes total internal reflection

Answer 1

Answer: (A)

We have, $\quad \mu_{1}=1.20+\frac{0.8 \times 10^{-14}}{\left(400 \times 10^{-9}\right)^{2}}$
or $\mu_{1}=1.20+\frac{0.8 \times 10^{-14}}{400 \times 400 \times 10^{-18}}$
or $\mu_{1}=1.20+\frac{0.8}{16}$
or $\mu_{1}=1.20+0.05$
or $\mu_{1}=1.25$
or $\sin i_{c}=\frac{1}{1.25}=0.8$
$\Rightarrow i_{c}=53.13^{\circ}$
Again, $\mu_{2}=1.20+\frac{0.8 \times 10^{-14}}{\left(500 \times 10^{-9}\right)^{2}}$

or $\mu_{2}=1.20+\frac{0.8}{25}$
or $ \mu_{2}=1.20+0.32$
or $\mu_{2}=1.232$
Or $\sin i_{c}=\frac{1}{1.232}=0.81$
$\Rightarrow i_{c}=\sin ^{-1} 0.81=54.26^{\circ}$
Now, $\sin\, \theta=0.8$ or $\theta=53.13^{\circ}$
This angle is clearly greater than critical angle corresponding to wavelength $400\, \,nm$.
So, light of $400\, \,nm$ wavelength undergoes total internal reflection.