Question

Consider a configuration of $n$ identical units, each consisting of three layers The first layer is a column of air of height $h=\frac{1}{3} cm$, and the second and third layers are of equal thickness $d=\frac{\sqrt{3}-1}{2} cm$, and refractive indices $\mu_1=\sqrt{\frac{3}{2}}$ and $\mu_2=\sqrt{3}$, respectively. A light source $O$ is placed on the top of the first unit, as shown in the figure. A ray of light from $O$ is incident on the second layer of the first unit at an angle of $\theta=60^{\circ}$ to the normal. For a specific value of $n$, the ray of light emerges from the bottom of the configuration at a distance $l=\frac{8}{\sqrt{3}} cm$, as shown in the figure The value of $n$ is ______

Answer 1

Correct Answer: 4

We start by noting that:

\( 1 \sin 60^\circ = \frac{\sqrt{3}}{2} r \)

Thus, \( r = 45^\circ \).

Next, we compute:

\( \frac{\sqrt{3}}{2} \sin 45^\circ = \sqrt{3} \sin r_2 \)

which gives \( r_2 = 30^\circ \).

Finally, we solve for:

\(\left( \frac{1}{\sqrt{3}} + \frac{\sqrt{3} - 1}{2 \sqrt{3}} + \frac{\sqrt{3} - 1}{2 \sqrt{3}} \right) \times n\)

\(= \frac{8}{\sqrt{3}} \quad \Rightarrow \quad n = 4.\)