Why does the $4000\ \text{\AA}$ wavelength undergo TIR and not the $5000\ \text{\AA}$ one?

The refractive index $\mu$ is inversely proportional to the square of the wavelength. A smaller wavelength ($4000\ \text{\AA}$) has a higher refractive index, which means a smaller critical angle. Thus, it satisfies the TIR condition before the larger wavelength.

How do we know the angle of incidence at face AC is $\theta$?

The rays enter face AB normally, so they travel straight through. Face AB is vertical, and face BC is horizontal. The normal to face AC makes an angle $\theta$ with the horizontal. Therefore, the horizontal rays strike face AC at an angle $\theta$ with the normal.

JEE Advanced 1991

LEVELJEE Advanced

Animated Solution for Physics - Optics: Two parallel beams of light $P$ and $Q$ (separation $d$ ) containing radiations of wavelengths $4000 \overset{˚}{A}$ and $5000 \overset{˚}{A}$ (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in figure. The refractive index of the prism as a function of wavelength is given by the relation, $μ (λ) = 1.20 + \frac{b}{λ ^{2}}$ where $λ$ is in $\overset{˚}{A}$ and $b$ is positive constant. The value of $b$ is such that the condition for total reflection at the face $A C$ is just satisfied for one wavelength and is not satisfied for the other. (a) Find the value of $b$ . (b) Find the deviation of the beams transmitted through the face $A C$ . (c) A convergent lens is used to bring these transmitted beams into focus. If the intensities of the upper and the lower beams immediately after transmission from the face $A C$ , are $4 I$ and $I$ respectively, find the resultant intensity at the focus.

Q: How is the resultant intensity calculated for coherent sources?

For coherent sources with a phase difference $\Delta \phi$, the resultant intensity is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$. Since the path difference is zero, $\Delta \phi = 0$, and the formula simplifies to $I = (\sqrt{I_1} + \sqrt{I_2})^2$.

Visualized Solution

\text{Geometry of the Prism}

The prism has a vertical face $A B$ and a horizontal face $B C$ .
The angle at $A$ is $θ$ , with $sin θ = 0.8$ .
The rays $P$ and $Q$ are incident normally on $A B$ , so they pass undeviated into the prism and strike the face $A C$ .
The angle of incidence at the face $A C$ is exactly $θ$ .

\text{Condition for Total Internal Reflection}

For Total Internal Reflection (TIR) to occur at face $A C$ , the angle of incidence must be greater than or equal to the critical angle $θ_{c}$ .
$i \geq θ_{c} ⟹ sin θ \geq sin θ_{c} = \frac{1}{μ}$
Therefore, TIR requires $μ \geq \frac{1}{s i n θ} = \frac{1}{0.8} = 1.25$ .

\text{Wavelength Dependence of TIR}

The refractive index is given by $μ (λ) = 1.20 + \frac{b}{λ ^{2}}$ .
Since $b > 0$ , $μ$ is larger for smaller wavelengths. Thus, $μ_{4000} > μ_{5000}$ .
The problem states TIR is just satisfied for one wavelength and not the other.
This means TIR occurs for $4000 \overset{˚}{A}$ (larger $μ$ ) but not for $5000 \overset{˚}{A}$ .
So, for $4000 \overset{˚}{A}$ , $μ_{4000} = 1.25$ .

\text{Calculating the Constant } b

Substitute $μ_{4000} = 1.25$ into the dispersion relation:
$1.25 = 1.20 + \frac{b}{( 4000 ) ^{2}}$
$\frac{b}{16 \times 1 0 ^{6}} = 0.05$
$b = 0.05 \times 16 \times 1 0^{6} = 8 \times 1 0^{5} \overset{˚}{A}^{2}$ .

\text{Deviation for } 4000\ \text{\AA}

For $4000 \overset{˚}{A}$ , the TIR condition is just satisfied, meaning the refracted ray grazes the surface $A C$ ( $r = 9 0^{\circ}$ ).
The deviation $δ$ is the angle between the incident ray direction and the emergent ray.
$δ_{4000} = 9 0^{\circ} - i = 9 0^{\circ} - θ$ .
Since $sin θ = 0.8$ , $θ \approx 53.1 3^{\circ}$ .
$δ_{4000} = 9 0^{\circ} - 53.1 3^{\circ} = 36.8 7^{\circ} \approx 3 7^{\circ}$ .

\text{Deviation for } 5000\ \text{\AA}

First, find $μ$ for $5000 \overset{˚}{A}$ :
$μ_{5000} = 1.20 + \frac{8 \times 1 0 ^{5}}{( 5000 ) ^{2}} = 1.20 + \frac{8 \times 1 0 ^{5}}{25 \times 1 0 ^{6}} = 1.20 + 0.032 = 1.232$ .
Apply Snell's Law at face $A C$ : $μ_{5000} sin θ = 1 \cdot sin r$ .
$1.232 \times 0.8 = sin r ⟹ sin r = 0.9856$ .
$r = sin^{- 1} (0.9856) \approx 80.2 6^{\circ}$ .
Deviation $δ_{5000} = r - i = 80.2 6^{\circ} - 53.1 3^{\circ} = 27.1 3^{\circ}$ .

\text{Path Difference of Transmitted Beams}

The $5000 \overset{˚}{A}$ beams emerge from the prism parallel to each other.
Because they originate from the same plane wavefront and undergo the same parallel refraction, they form a new plane wavefront upon emergence.
A convergent lens focuses parallel rays to its focal point with zero additional path difference.
Therefore, the net phase difference $Δ ϕ$ between the two beams at the focus is $0$ .

\text{Resultant Intensity at the Focus}

The two beams interfere constructively at the focus since $Δ ϕ = 0$ .
The resultant intensity $I_{res}$ is given by:
$I_{res} = (I_{1} + I_{2})^{2}$
Given $I_{1} = 4 I$ and $I_{2} = I$ :
$I_{res} = (4 I + I)^{2} = (2 I + I)^{2} = (3 I)^{2} = 9 I$ .

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The Sigma Insight: Refraction and Dispersion through Prism

Solution Diagram

Analyzing the Setup

Imagine two parallel beams of light,

P

and

Q

, striking a prism. They hit the first face normally, marching straight in without any deviation. But the real magic happens at the second face,

A C

. Here, the rays strike at an angle

θ

. The prism isn't just any ordinary glass; its refractive index changes with wavelength, a phenomenon known as dispersion.

The Master Equation for TIR

Total Internal Reflection (TIR) is a strict bouncer. It only lets rays reflect internally if their angle of incidence exceeds the critical angle. The critical angle depends on the refractive index. Since

μ

is larger for the

4000 \overset{˚}{A}

light, its critical angle is smaller. The problem tells us TIR is just satisfied for one wavelength. Naturally, the

4000 \overset{˚}{A}

light hits the critical angle first!

Final Calculation

By setting the angle of incidence

θ

equal to the critical angle for

4000 \overset{˚}{A}

, we elegantly solve for the constant

b

. For the

5000 \overset{˚}{A}

light, we apply Snell's Law to find its angle of emergence and subsequent deviation.

Finally, the grand finale: the

5000 \overset{˚}{A}

beams emerge parallel, forming a plane wavefront. When a lens focuses them, they meet with zero path difference, interfering constructively to give a brilliant resultant intensity of

9 I

Common Traps & FAQs

Why does the

4000 \overset{˚}{A}

wavelength undergo TIR and not the

5000 \overset{˚}{A}

one?

The refractive index

μ

is inversely proportional to the square of the wavelength. A smaller wavelength (

4000 \overset{˚}{A}

) has a higher refractive index, which means a smaller critical angle. Thus, it satisfies the TIR condition before the larger wavelength.

How do we know the angle of incidence at face AC is

θ

The rays enter face AB normally, so they travel straight through. Face AB is vertical, and face BC is horizontal. The normal to face AC makes an angle

θ

with the horizontal. Therefore, the horizontal rays strike face AC at an angle

θ

with the normal.

Why is the path difference zero for the beams focused by the lens?

The beams emerge parallel from the prism, meaning they form a plane wavefront. A convergent lens focuses all rays from a plane wavefront to a single point (the focus) such that the optical path length for all rays is identical.

What does "just satisfied" mean for the TIR condition?

It means the angle of incidence is exactly equal to the critical angle. In this case, the refracted ray grazes the boundary, making the angle of refraction

9 0^{\circ}

How is the resultant intensity calculated for coherent sources?

For coherent sources with a phase difference

Δ ϕ

, the resultant intensity is

I = I_{1} + I_{2} + 2 I_{1} I_{2} cos (Δ ϕ)

. Since the path difference is zero,

Δ ϕ = 0

, and the formula simplifies to

I = (I_{1} + I_{2})^{2}