Question

A prism of refractive index $n_1$ and another prism of refractive index $n_2$ are stuck together (as shown in the figure). $n_1$ and $n_2$ depend on $\lambda$, the wavelength of light, according to the relation $n_1 = 1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2}$ and $n_2 = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}$. The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be _________ nm. 

 

Hint:

The condition for light to pass undeviated from one medium to another is that their refractive indices must be equal. This problem combines this optical principle with Cauchy's relation for dispersion (refractive index depending on wavelength).

Answer 1

Correct Answer: 600

For a ray of light to pass through the interface BC without bending (without deviation), the refractive indices of the two media must be equal.
According to Snell's Law, $n_1 \sin i = n_2 \sin r$. For no bending, the angle of refraction r must be equal to the angle of incidence i. This is only possible if $n_1 = n_2$.
We are given the relations for $n_1$ and $n_2$ as a function of wavelength $\lambda$. We set them equal to each other:
$n_1 = n_2$
$1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2} = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}$
Rearrange the equation to solve for $\lambda^2$:
$\frac{10.8 \times 10^{-14}}{\lambda^2} - \frac{1.8 \times 10^{-14}}{\lambda^2} = 1.45 - 1.2$
$\frac{(10.8 - 1.8) \times 10^{-14}}{\lambda^2} = 0.25$
$\frac{9.0 \times 10^{-14}}{\lambda^2} = 0.25$
$\lambda^2 = \frac{9.0 \times 10^{-14}}{0.25} = \frac{9.0 \times 10^{-14}}{1/4} = 36 \times 10^{-14}$ m$^2$.
Now, take the square root to find the wavelength $\lambda$:
$\lambda = \sqrt{36 \times 10^{-14}} = 6 \times 10^{-7}$ m.
The question asks for the answer in nanometers (nm).
$\lambda = 6 \times 10^{-7} \text{ m} = 600 \times 10^{-9} \text{ m} = 600$ nm.