Question

A transparent container contains layers of three immiscible transparent liquids \(A\), \(B\) and \(C\) of refractive indices \(n\), \(\dfrac{3n}{4}\) and \(\dfrac{2n}{3}\), respectively. A laser beam is incident at the interface between \(A\) and \(B\) at an angle \(\theta\) as shown in the figure. Prove that the beam does not enter region \(C\) at all for \[ \sin\theta>\frac{2}{3}. \]

Hint:

For problems involving several layers of liquids:
• Use Snell's law successively at each interface.
• Find the critical angle at the final interface.
• Apply the condition for total internal reflection: \[ i>C \qquad\text{or}\qquad \sin i>\sin C. \] In this problem, \[ \boxed{ \sin r=\frac43\sin\theta } \qquad\text{and}\qquad \boxed{ \sin C=\frac89 } \] which finally gives \[ \boxed{ \sin\theta>\frac23. } \]

Answer 1

Concept: As the light ray travels from one medium to another, its direction changes according to Snell's law: \[ n_1\sin i=n_2\sin r. \] The three liquids have refractive indices \[ \mu_A=n,\qquad \mu_B=\frac{3n}{4},\qquad \mu_C=\frac{2n}{3}. \] Since \[ n>\frac{3n}{4}>\frac{2n}{3}, \] each successive medium is optically rarer than the previous one. Consequently, the ray bends away from the normal at every interface. To prove that the beam does not enter region \(C\), we must show that the ray undergoes total internal reflection at the interface between liquids \(B\) and \(C\).

Step 1: Apply Snell's law at the interface between liquids \(A\) and \(B\).
Let the angle of refraction in liquid \(B\) be \(r\). Applying Snell's law, \[ n\sin\theta = \frac{3n}{4}\sin r. \] Cancelling \(n\) from both sides, \[ \sin r = \frac{4}{3}\sin\theta. \] Therefore, \[ \boxed{ \sin r=\frac{4}{3}\sin\theta } \]

Step 2: Find the critical angle for the interface between liquids \(B\) and \(C\).
The refractive index of liquid \(B\) is \[ \mu_B=\frac{3n}{4} \] and that of liquid \(C\) is \[ \mu_C=\frac{2n}{3}. \] If \(C\) is the critical angle for the interface \(B-C\), then \[ \sin C = \frac{\mu_C}{\mu_B}. \] Substituting the values, \[ \sin C = \frac{\frac{2n}{3}} {\frac{3n}{4}}. \] \[ \sin C = \frac{2n}{3}\times\frac{4}{3n} = \frac{8}{9}. \] Hence, \[ \boxed{ \sin C=\frac{8}{9} } \]

Step 3: Determine the condition for total internal reflection.
For the ray not to enter liquid \(C\), the angle of incidence at the \(B-C\) interface must be greater than the critical angle. That is, \[ r>C. \] Since the sine function is increasing in the range \(0^\circ\) to \(90^\circ\), \[ \sin r>\sin C. \] Substituting the values of \(\sin r\) and \(\sin C\), \[ \frac{4}{3}\sin\theta > \frac{8}{9}. \] Multiplying both sides by \(\dfrac34\), \[ \sin\theta > \frac{8}{9}\times\frac34. \] Therefore, \[ \boxed{ \sin\theta>\frac23 } \] Thus, whenever \[ \boxed{ \sin\theta>\frac23, } \] the angle of incidence at the interface between liquids \(B\) and \(C\) exceeds the critical angle and the ray undergoes total internal reflection. Hence, the laser beam does not enter region \(C\) at all. \[ \boxed{ \text{For } \sin\theta>\frac23,\ \text{the beam is totally internally reflected in liquid }B\text{ and never enters region }C. } \]