A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers

Q: Why is the refracted ray parallel to the base at minimum deviation?

At minimum deviation, the light path is perfectly symmetrical. If you reverse the light ray, it must retrace its path with the same deviation. This symmetry forces the ray inside the prism to be parallel to the base (for an isosceles or equilateral prism).

Q: Can I use the prism formula directly instead of Snell's Law?

Absolutely! The prism formula $n = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$ is derived from Snell's Law and the symmetry condition. You can use it to find $\delta_m$ first, and then use $i = (A+\delta_m)/2$. Both methods are perfectly valid and yield the same result.

Q: What if the prism was not equilateral?

If the prism was not equilateral, the prism angle $A$ would be different. The condition $r = A/2$ still holds for minimum deviation, but you would substitute the specific value of $A$ for that prism.

Q: Why did we take $A = 60^\circ$?

The problem explicitly states it is an 'equilateral triangular prism'. In an equilateral triangle, all internal angles are exactly $60^\circ$.

Q: Does the wavelength of the monochromatic light matter here?

The refractive index $n$ depends on the wavelength of light (dispersion). However, the problem already provides the specific refractive index ($n = \sqrt{3}$) for this particular monochromatic light, so we don't need to worry about its exact wavelength.

Question

A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $\sqrt{3}$, then the angle of incidence is

Options:

(A) $45^\circ$

(B) $90^\circ$

(C) $60^\circ$

(D) $30^\circ$

Sigma Percentile · Accepted Answer

Have you ever wondered what happens when light passes through a prism in the most perfectly symmetrical way possible? This special condition is known as **minimum deviation**, and it unlocks a beautiful geometric harmony within the prism. Let's dive into this fascinating problem and uncover the angle of incidence!

### The Magic of Symmetry

When a monochromatic light ray suffers minimum deviation through a prism, it doesn't just take any random path. It chooses the path of perfect symmetry. Imagine the light ray entering the prism, traveling through it, and exiting. At minimum deviation, the ray inside the prism becomes exactly parallel to the base of the prism (assuming it's an isosceles or equilateral prism).

Because of this beautiful symmetry, the angle of refraction $r$ at the first interface is simply half of the prism's apex angle $A$.

$$ r = \frac{A}{2} $$

Since we are dealing with an **equilateral triangular prism**, we know that all its internal angles are $60^\circ$. Therefore, our prism angle $A$ is $60^\circ$.

$$ r = \frac{60^\circ}{2} = 30^\circ $$

### Bridging the Worlds with Snell's Law

Now that we know the angle of refraction inside the prism, we need to find the angle of incidence $i$ outside the prism. To connect these two worlds—the air outside and the glass inside—we call upon our trusty tool: **Snell's Law**.

Snell's Law states that the product of the refractive index and the sine of the angle is constant across an interface:

$$ n_1 \sin i = n_2 \sin r $$

Here, the light is coming from air (where $n_1 \approx 1$) and entering the prism, which has a given refractive index of $n_2 = \sqrt{3}$. Let's substitute our known values into the equation:

$$ 1 \cdot \sin i = \sqrt{3} \cdot \sin 30^\circ $$

### The Final Calculation

We are almost there! From standard trigonometry, we know that the sine of $30^\circ$ is exactly $\frac{1}{2}$. Let's plug that in:

$$ \sin i = \sqrt{3} \cdot \left(\frac{1}{2}\right) $$

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

Now, we just need to ask ourselves: which angle has a sine value of $\frac{\sqrt{3}}{2}$? If you recall your trigonometric tables, the answer is $60^\circ$.

$$ i = 60^\circ $$

And there we have it! For the light ray to experience minimum deviation through this equilateral prism, it must strike the surface at an angle of incidence of **$60^\circ$**. The elegance of this problem lies in how physical phenomena like refraction perfectly align with geometric symmetry.

Animated Solution for Physics - Optics: A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $3$ , then the angle of incidence is

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Visualized Solution

$The Prism Setup$

$Condition for Minimum Deviation$

$Calculating Angle of Refraction$

$Snell’s Law at First Interface$

$Substituting Values$

$Evaluating the Expression$

$Finding the Angle of Incidence$

$Food for Thought$

The Sigma Insight: Refraction and Dispersion through Prism

The Magic of Symmetry

Bridging the Worlds with Snell's Law

The Final Calculation

Similar Questions

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Visualized Solution

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

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

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

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

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

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

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

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