Cross-section view of a prism is the equilateral triangle ABC in the figure. The minimum deviation i

Q: Why did we assume the angle of incidence is equal to the prism angle?

We didn't assume it; the problem explicitly states: 'The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle.' This is a specific condition given for this particular setup.

Q: How do we know that $r = A/2$?

At the angle of minimum deviation, the light ray passes symmetrically through the prism. This symmetry means the refracted ray is parallel to the base, and the angle of refraction at the first face is exactly half of the prism's apex angle.

Q: What exactly is an optical path?

The optical path is the distance light would travel in a perfect vacuum in the same amount of time it takes to travel through a specific medium. It is calculated by multiplying the geometric distance traveled in the medium by the medium's refractive index ($\mu \cdot d$).

Q: Why did we use $\cos 30^\circ$ to find the length of AP?

In the equilateral triangle $ABC$, the line $AP$ drops from the apex to the midpoint of the base, bisecting the $60^\circ$ angle at $A$ into two $30^\circ$ angles. This forms a right-angled triangle $ABP$. Using trigonometry, the adjacent side $AP$ is the hypotenuse $AB$ multiplied by $\cos 30^\circ$.

Q: Why did we have to convert the distance AP from centimeters to meters?

The speed of light $c$ is given in standard SI units as $3 \times 10^8$ meters per second. To ensure our final time is correctly calculated in seconds, all distances in the equation must also be in meters.

Question

Cross-section view of a prism is the equilateral triangle $ABC$ in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from $P$ (mid-point of $BC$) to $A$ is ..... $	imes 10^{-10}$ s. (Given, speed of light in vacuum $= 3 	imes 10^8$ m/s and $\cos 30^\circ = \frac{\sqrt{3}}{2}$)

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine an equilateral prism with a side of $10	ext{ cm}$. We are tasked with finding the time it takes for light to travel straight up from point $P$, the midpoint of the base, to the apex $A$.

Before we can calculate the time, we need to understand the optical properties of the medium the light is traveling through. This requires us to find the refractive index of the prism.

### The Magic of Minimum Deviation

The problem provides a crucial piece of information: the prism is at **minimum deviation**, and this occurs when the angle of incidence $i$ is equal to the prism angle $A$.

Since the prism is an equilateral triangle, we know that the prism angle $A = 60^\circ$. Therefore, the angle of incidence is also $i = 60^\circ$.

At the position of minimum deviation, the refracted ray travels perfectly parallel to the base of the prism. Because of this symmetry, the angle of refraction $r$ at the first surface is exactly half of the prism angle.

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

### Unlocking the Refractive Index

Now that we have both the angle of incidence and the angle of refraction, we can invoke **Snell's Law** to find the refractive index $\mu$ of the prism material.

$$\mu = \frac{\sin i}{\sin r}$$

Substituting our known angles into the equation:

$$\mu = \frac{\sin 60^\circ}{\sin 30^\circ}$$

We know the standard trigonometric values: $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$. Plugging these in:

$$\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$$

### The Concept of Optical Path

To find the time $t$ taken by light to travel from $P$ to $A$, we use the concept of the **optical path**. The optical path is the equivalent distance light would travel in a vacuum in the same amount of time it takes to travel through a medium.

The time taken is simply the optical path divided by the speed of light in a vacuum, $c$:

$$t = \frac{	ext{Optical Path}}{c} = \frac{\mu \cdot AP}{c}$$

### Geometry Meets Physics

We need to calculate the geometric distance $AP$. Looking at the right-angled triangle $\Delta ABP$, the line $AP$ bisects the $60^\circ$ angle at $A$, making $\angle BAP = 30^\circ$.

Using basic trigonometry:

$$AP = AB \cos 30^\circ$$

Given that the side of the equilateral triangle $AB = 10	ext{ cm}$:

$$AP = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}	ext{ cm}$$

To maintain dimensional consistency with the speed of light, we must convert this distance into meters:

$$AP = 5\sqrt{3} 	imes 10^{-2}	ext{ m}$$

### The Final Countdown

We now have all the pieces of the puzzle. Let's substitute them into our time equation:

$$t = \frac{\sqrt{3} \cdot (5\sqrt{3} 	imes 10^{-2})}{3 	imes 10^8}$$

Multiplying the numerators, $\sqrt{3} \cdot \sqrt{3} = 3$. This perfectly cancels out the $3$ in the denominator:

$$t = \frac{15 	imes 10^{-2}}{3 	imes 10^8}$$

$$t = 5 	imes 10^{-10}	ext{ s}$$

Comparing this result with the format given in the question ($..... 	imes 10^{-10}	ext{ s}$), the missing integer is exactly **$5$**.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$M inim u m D e v ia t i o n C o n d i t i o n$

$A n g l e o f R e f r a c t i o n$

$S n e l l^{'} s L a w S e t u p$

$R e f r a c t i v e I n d e x C a l c u l a t i o n$

$O pt i c a l P a t h C o n ce pt$

$G eo m e t r i c D i s t an ce A P$

$S u b s t i t u t i o n$

$F ina l C a l c u l a t i o n$

$F ina l A n s w er$

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

The Sigma Insight: Refraction and Dispersion through Prism

Visualizing the Setup

The Magic of Minimum Deviation

Unlocking the Refractive Index

The Concept of Optical Path

Geometry Meets Physics

The Final Countdown

Similar Questions

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A ray of light is incident normally on one of the faces of a prism of apex angle $3 0^{\circ}$ and refractive index $2$ . The angle of deviation of the ray is ...... degrees.

Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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

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

.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