Consider a tank made of glass (refractive index is 1.5) with a thick bottom. It is filled with a liq

Q: Why do we use the critical angle $i_C$ here?

The critical angle $i_C$ represents the absolute maximum possible angle of refraction when light enters from a rarer medium (air) to a denser medium (liquid). If even this maximum angle cannot reach Brewster's angle, then no other angle can.

Q: Why is the condition $i_C i_B$?

If $i_C > i_B$, it means the range of possible refraction angles includes $i_B$. The light could hit the interface exactly at $i_B$, causing complete polarization. We want to avoid this, so the maximum angle $i_C$ must fall short of $i_B$.

Q: How did we get $\sin(i_B)$ from $\tan(i_B)$?

By constructing a right-angled triangle. If $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1.5}{\mu}$, then the hypotenuse is $\sqrt{1.5^2 + \mu^2}$. Therefore, $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1.5}{\sqrt{1.5^2 + \mu^2}}$.

Q: Why is $\sin(i_C) = \frac{1}{\mu}$?

We apply Snell's law at the air-liquid interface for grazing incidence (where the incident angle in air is $90^\circ$). The equation is $1 \cdot \sin(90^\circ) = \mu \cdot \sin(i_C)$, which simplifies to $\sin(i_C) = \frac{1}{\mu}$.

Q: Does the thickness of the glass bottom matter?

No, polarization by reflection depends entirely on the angle of incidence and the refractive indices of the two media exactly at the boundary. The thickness of the glass does not alter Brewster's angle.

Question

Consider a tank made of glass (refractive index is $1.5$) with a thick bottom. It is filled with a liquid of refractive index $\mu$. A student finds that, irrespective of what the incident angle $i$ (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarised. For this to happen, the minimum value of $\mu$ is

Options:

(A) $\frac{3}{\sqrt{5}}$

(B) $\frac{5}{\sqrt{3}}$

(C) $\frac{4}{3}$

(D) $\frac{5}{3}$

Sigma Percentile · Accepted Answer

### Unlocking the Secrets of Polarization and Refraction Imagine you are looking at a glass tank filled with a mysterious liquid. The setup is simple: air on top, a liquid of unknown refractive index $\mu$ in the middle, and a thick glass bottom with a refractive index of $1.5$. A beam of light enters the liquid from the air, bends, and then strikes the glass bottom, reflecting back up. This problem presents a fascinating constraint: no matter what angle the light enters the liquid, the light reflected from the glass bottom is **never completely polarized**. Let's decode what this means physically and mathematically. ### The Condition for Polarization When light reflects off a boundary between two transparent media, it can become completely plane-polarized. However, this only happens at one specific angle of incidence, known as **Brewster's angle** ($i_B$). According to Brewster's Law, the tangent of this angle is equal to the ratio of the refractive indices of the two media. For our liquid-glass interface, this means: $$ an(i_B) = \frac{\mu_{glass}}{\mu_{liquid}} = \frac{1.5}{\mu}$$ If the light ray inside the liquid strikes the glass at exactly this angle $i_B$, the reflected ray will be perfectly polarized. ### The Constraint: Why it Never Polarizes The problem states that complete polarization *never* happens. This implies a physical impossibility: the light ray traveling through the liquid can never reach an angle steep enough to equal Brewster's angle. Let's trace the light's journey. It enters from air (a rarer medium) into the liquid (a denser medium). It refracts at an angle $r$. By simple geometry, this angle $r$ is also the angle of incidence when the ray hits the glass bottom. For the light to never polarize, the maximum possible value of $r$ must be strictly less than $i_B$. But what is the maximum value of $r$? It occurs when the light enters the liquid at a grazing angle (an incident angle of $90^\circ$ in air). This maximum angle of refraction is exactly the **critical angle** ($i_C$) for the air-liquid interface. Therefore, our master constraint is: $$i_C < i_B$$ ### Mathematical Formulation Since the sine function is strictly increasing in the first quadrant, we can rewrite our constraint as: $$\sin(i_C) < \sin(i_B)$$ Let's find expressions for both sides. First, applying Snell's Law at the air-liquid interface for grazing incidence ($i = 90^\circ$): $$1 \cdot \sin(90^\circ) = \mu \cdot \sin(i_C)$$ $$\sin(i_C) = \frac{1}{\mu}$$ Next, we need $\sin(i_B)$. We already know $ an(i_B) = \frac{1.5}{\mu}$. Imagine a right-angled triangle where the opposite side is $1.5$ and the adjacent side is $\mu$. The hypotenuse is $\sqrt{1.5^2 + \mu^2}$. Therefore: $$\sin(i_B) = \frac{1.5}{\sqrt{1.5^2 + \mu^2}}$$ ### The Final Calculation Now, we substitute these into our inequality: $$\frac{1}{\mu} < \frac{1.5}{\sqrt{1.5^2 + \mu^2}}$$ To solve this, we square both sides to eliminate the square root: $$\frac{1}{\mu^2} < \frac{2.25}{2.25 + \mu^2}$$ Cross-multiplying gives us a simple algebraic inequality: $$2.25 + \mu^2 < 2.25 \mu^2$$ Rearranging the terms to isolate $\mu$: $$2.25 < 1.25 \mu^2$$ $$\mu^2 > \frac{2.25}{1.25} = \frac{9}{5}$$ Taking the square root of both sides, we arrive at our final answer: $$\mu > \frac{3}{\sqrt{5}}$$ Thus, for the reflected light to never be completely polarized, the refractive index of the liquid must be strictly greater than $\frac{3}{\sqrt{5}}$. The minimum bounding value is exactly $\frac{3}{\sqrt{5}}$.

Select Answer:

Visualized Solution

Visualizing the Setup

Tracing the Light Ray

Reflection at the Bottom

Brewster's Law

The Constraint

Maximum Angle of Refraction

Setting up the Inequality

Calculating $sin (i_{C})$

Calculating $tan (i_{B})$

Converting $tan (i_{B})$ to $sin (i_{B})$

Substituting into the Inequality

Solving the Inequality

Final Algebraic Steps

Conclusion

The Sigma Insight: Polarization

Unlocking the Secrets of Polarization and Refraction

The Condition for Polarization

The Constraint

Mathematical Formulation

The Final Calculation

Similar Questions

The angle of incidence at which reflected light is totally polarised for reflection from air to glass (refractive index $n$ ), is

When an unpolarised light of intensity $I_{0}$ is incident on a polarising sheet, the intensity of the light which does not get transmitted is

A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $4 5^{\circ}$ relative to that of $A$ . The intensity of the emergent light is

Select Answer:

Visualized Solution

Visualizing the Setup

Tracing the Light Ray

Reflection at the Bottom

Brewster's Law

The Constraint

Maximum Angle of Refraction

Setting up the Inequality

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(iC​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(iB​)

Converting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(iB​) to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(iB​)

Substituting into the Inequality

Solving the Inequality

Final Algebraic Steps

Conclusion

The Sigma Insight: Polarization

Unlocking the Secrets of Polarization and Refraction

The Condition for Polarization

The Constraint

Mathematical Formulation

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle of incidence at which reflected light is totally polarised for reflection from air to glass (refractive index n), is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } When an unpolarised light of intensity I0​ is incident on a polarising sheet, the intensity of the light which does not get transmitted is

Calculating $sin (i_{C})$

Calculating $tan (i_{B})$

Converting $tan (i_{B})$ to $sin (i_{B})$

The angle of incidence at which reflected light is totally polarised for reflection from air to glass (refractive index $n$ ), is

When an unpolarised light of intensity $I_{0}$ is incident on a polarising sheet, the intensity of the light which does not get transmitted is