A prism of refracting angle 30 degrees is coated with a thin film of transparent material of refract

Q: Why is the angle of emergence $0^{\circ}$ and not $90^{\circ}$?

The angle of emergence is measured from the normal. Since the ray strikes the face AC perpendicularly, it travels exactly along the normal line. Thus, the angle it makes with the normal is $0^{\circ}$.

Q: Why do we use $2\mu t = n\lambda$ for constructive interference in transmission? Shouldn't it be $(n - 1/2)\lambda$?

The condition heavily depends on the phase changes at the boundaries. Here, light reflects from the film-air boundary ($2.2$ to $1.0$) and the film-prism boundary ($2.2$ to $\sqrt{3}$). Both are reflections from a rarer medium, so neither introduces a $\pi$ phase shift. Thus, the net phase shift is zero, and the condition for constructive interference remains $2\mu t = n\lambda$.

Q: What if the ray didn't hit the film normally?

If the ray hit the film at an angle of refraction $r_f$, the path difference would be $2\mu t \cos r_f$. Because it hits normally, $r_f = 0^{\circ}$ and $\cos 0^{\circ} = 1$, simplifying the path difference to just $2\mu t$.

Q: Why do we take $n=1$ for minimum thickness?

The thickness $t$ is directly proportional to the integer $n$. To find the minimum non-zero thickness that satisfies the constructive interference condition, we must choose the smallest positive integer, which is $n=1$.

Q: Does the refractive index of the prism affect the interference condition?

Yes, indirectly! It determines whether a phase shift occurs at the film-prism boundary. If the prism had a higher refractive index than the film (e.g., $\mu_p = 2.5$), a $\pi$ phase shift would occur upon reflection, completely flipping the condition for constructive interference.

Question

A prism of refracting angle $30^{\circ}$ is coated with a thin film of transparent material of refractive index $2.2$ on face $AC$ of the prism. A light of wavelength $6600 	ext{ \AA}$ is incident on face $AB$ such that angle of incidence is $60^{\circ}$. Find
(a) the angle of emergence and
(b) the minimum value of thickness of the coated film on the face $AC$ for which the light emerging from the face has maximum intensity. (Given refractive index of the material of the prism is $\sqrt{3}$)

Sigma Percentile · Accepted Answer

This problem is a beautiful fusion of two classic optics concepts: the geometry of a prism and the wave nature of light in thin films. Let's embark on this journey step by step.

### Analyzing the Setup

Imagine a prism with a refracting angle $A = 30^{\circ}$. A light ray strikes the first face, $AB$, at an angle of incidence $i_1 = 60^{\circ}$. The second face, $AC$, is coated with a thin film of refractive index $\mu_f = 2.2$. Our first mission is to trace the path of this ray through the prism to see how it interacts with the film.

### Tracing the Ray Through the Prism

When the light enters the prism from the air, it bends towards the normal. We apply Snell's Law at the first interface:

$$1 \cdot \sin 60^{\circ} = \mu_p \sin r_1$$

Given the prism's refractive index $\mu_p = \sqrt{3}$, we substitute the values:

$$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1$$

$$\sin r_1 = \frac{1}{2} \implies r_1 = 30^{\circ}$$

Now, we use the fundamental geometric property of a prism, which relates the internal angles of refraction to the prism angle:

$$r_1 + r_2 = A$$

Substituting our known values:

$$30^{\circ} + r_2 = 30^{\circ} \implies r_2 = 0^{\circ}$$

What does $r_2 = 0^{\circ}$ mean physically? It means the ray strikes the second face, $AC$, perfectly perpendicular to the surface! Because it hits normally, it passes straight through the film and emerges into the air without any deviation. Thus, the angle of emergence is **$0^{\circ}$**.

### The Thin Film Interference

Now, let's zoom into the thin film on face $AC$. As the light passes through, it undergoes multiple internal reflections between the film-air boundary and the film-prism boundary. We want the transmitted light to have maximum intensity, which means the directly transmitted ray and the twice-reflected transmitted ray must undergo **constructive interference**.

Before writing the interference condition, we must carefully check for phase changes upon reflection.

1. **Reflection at the film-air boundary:** The light in the film ($\mu_f = 2.2$) reflects off the air ($\mu_a = 1.0$). Since it's reflecting from a rarer medium, there is **no phase change**.
2. **Reflection at the film-prism boundary:** The light in the film ($\mu_f = 2.2$) reflects off the prism ($\mu_p = \sqrt{3} \approx 1.732$). Again, it's reflecting from a rarer medium, so there is **no phase change**.

Since the net phase change from reflections is zero, the condition for constructive interference in transmission is simply that the optical path difference must be an integer multiple of the wavelength:

$$2 \mu_f t = n \lambda$$

### Final Calculation

We are asked for the **minimum** value of thickness $t$. To minimize $t$, we choose the smallest positive integer, $n = 1$. Rearranging our master equation:

$$t = \frac{\lambda}{2 \mu_f}$$

Substitute the given wavelength $\lambda = 6600 	ext{ \AA}$ and the film's refractive index $\mu_f = 2.2$:

$$t = \frac{6600}{2 	imes 2.2} = \frac{6600}{4.4}$$

$$t = 1500 	ext{ \AA}$$

The minimum thickness required for the film to maximize the transmitted light intensity is **$1500 	ext{ \AA}$**.

Visualized Solution

$A na l y z in g t h e S e t u p$

$S n e l l^{'} s L a w a t F a ce A B$

$C a l c u l a t in g A n g l e o f R e f r a c t i o n r_{1}$

$G eo m e t r y o f t h e P r i s m$

$A n g l e o f I n c i d e n ce a t F a ce A C$

$A n g l e o f E m er g e n ce$

$I n t er f er e n ce in t h e T hin F i l m$

$P ha se C han g es o n R e f l ec t i o n$

$C o n d i t i o n f or C o n s t r u c t i v e I n t er f er e n ce$

$C a l c u l a t in g M inim u m T hi c k n ess$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The Sigma Insight: Refraction and Dispersion through Prism

Analyzing the Setup

Tracing the Ray Through the Prism

The Thin Film Interference

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Snell′s Law at Face AB

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Angle of Refraction r1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometry of the Prism

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle of Incidence at Face AC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle of Emergence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Interference in the Thin Film

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Phase Changes on Reflection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Constructive Interference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Minimum Thickness

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward