In an experiment for determination of refractive index of glass of a prism by i-delta, plot, it was

Q: Why can't we just plug the given 40 degrees directly into the prism formula as delta m?

The prism formula specifically uses the *minimum* angle of deviation ($\delta_m$). The 40 degrees given is just the deviation for a specific incidence angle of 35 degrees. We have no proof that 40 degrees is the absolute minimum deviation this prism can produce.

Q: How did we know that sin(37 degrees) is 0.6?

This comes from the classic 3-4-5 right-angled triangle, which is a staple in physics problems. In a 3-4-5 triangle, the angle opposite the side of length 3 is approximately 37 degrees, making $\sin 37^\circ = 3/5 = 0.6$.

Q: Why did we test the smallest option (1.5) first instead of the largest?

Because the refractive index $\mu$ increases as the minimum deviation $\delta_m$ increases. By testing the smallest option, we found that even it required a minimum deviation ($54^\circ$) that was too large (greater than $40^\circ$). This immediately proved that all larger options (1.6, 1.7, 1.8) would be even more impossible.

Q: What if the calculated minimum deviation for mu = 1.5 was less than 40 degrees?

If the calculated $\delta_m$ was, say, $38^\circ$, then $\mu = 1.5$ would be a physically possible refractive index. We would then have to test the next higher option (1.6) to see if it also yields a valid $\delta_m$, continuing until we find the maximum possible value.

Q: If the actual refractive index is strictly less than 1.5, why is 1.5 the correct answer?

The question specifically asks for the option that is *closest* to the maximum possible value. Since the true maximum possible value (calculated using $\delta_m = 40^\circ$) is approximately 1.39, the option 1.5 is the closest one provided in the multiple-choice list.

Question

In an experiment for determination of refractive index of glass of a prism by $i-\delta$, plot, it was found that a ray incident at an angle $35^\circ$ suffers a deviation of $40^\circ$ and that it emerges at an angle $79^\circ$. In that case, which of the following is closest to the maximum possible value of the refractive index?

Options:

(A) 1.5

(B) 1.6

(C) 1.7

(D) 1.8

Sigma Percentile · Accepted Answer

This problem is a beautiful example of how physics requires us to think beyond just plugging numbers into a formula. It tests our conceptual understanding of the minimum deviation in a prism and how it acts as a boundary condition for the refractive index.

### Decoding the Prism's Geometry

Imagine a light ray striking a glass prism. We are given three crucial pieces of information: the angle of incidence $i = 35^\circ$, the angle of emergence $e = 79^\circ$, and the angle of deviation $\delta = 40^\circ$.

Before we can even think about the refractive index, we need to understand the physical shape of the prism, specifically its refracting angle $A$. The fundamental geometric relation for any ray passing through a prism is that the total deviation is the sum of the incidence and emergence angles, minus the prism angle:

$$\delta = i + e - A$$

By substituting our known values into this equation, we get:

$$40^\circ = 35^\circ + 79^\circ - A$$

Solving this simple linear equation reveals that the angle of the prism $A$ is exactly $74^\circ$.

### The Master Equation of Refraction

Now that we have the prism angle, we can connect it to the refractive index $\mu$. The standard prism formula relates the refractive index to the prism angle $A$ and the **minimum angle of deviation**, $\delta_m$:

$$\mu = \frac{\sin\left(\frac{A+\delta_m}{2}ight)}{\sin\left(\frac{A}{2}ight)}$$

### The Minimum Deviation Trap

Here is where many students fall into a trap. The $40^\circ$ deviation given in the problem is just *one specific* deviation that occurs when the incidence angle is $35^\circ$. It is **not necessarily** the minimum deviation.

By definition, the minimum deviation $\delta_m$ is the absolute lowest deviation a ray can experience while passing through that specific prism. Therefore, the true minimum deviation must be less than or equal to any observed deviation. This gives us a strict mathematical constraint:

$$\delta_m \le 40^\circ$$

### Proof by Contradiction

Since we are looking for the maximum possible value of the refractive index $\mu$, let's test the options provided, starting with the smallest one: $\mu = 1.5$. If we assume $\mu = 1.5$, we can calculate what the minimum deviation *would* be.

$$1.5 = \frac{\sin\left(\frac{74^\circ+\delta_m}{2}ight)}{\sin\left(\frac{74^\circ}{2}ight)}$$

The denominator simplifies to $\sin 37^\circ$, which is approximately $0.6$. Multiplying both sides by $0.6$ gives:

$$1.5 	imes 0.6 = \sin\left(37^\circ + \frac{\delta_m}{2}ight)$$

$$0.9 = \sin\left(37^\circ + \frac{\delta_m}{2}ight)$$

We know from standard trigonometric values that $\sin 64^\circ \approx 0.9$. Equating the angles inside the sine function:

$$37^\circ + \frac{\delta_m}{2} = 64^\circ$$

$$\frac{\delta_m}{2} = 27^\circ \implies \delta_m = 54^\circ$$

This is a massive contradiction! We calculated that if $\mu = 1.5$, the minimum deviation would be $54^\circ$. However, we already know that a ray passed through the prism with a deviation of only $40^\circ$. It is physically impossible for the minimum deviation to be $54^\circ$ if a $40^\circ$ deviation exists.

Because the actual minimum deviation must be less than or equal to $40^\circ$, the true refractive index must be strictly less than $1.5$. Looking at our options ($1.5, 1.6, 1.7, 1.8$), the value $1.5$ is the closest maximum possible value. Any higher refractive index would require an even larger, more impossible minimum deviation.

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Visualized Solution

$Problem Statement$

$Prism Angle Formula$

$Calculating Prism Angle$

$Refractive Index Formula$

$The Minimum Deviation Catch$

$Testing the Options$

$Solving for δ_{m}$

$The Contradiction$

$Conclusion$

The Sigma Insight: Refraction and Dispersion through Prism

Decoding the Prism's Geometry

The Master Equation of Refraction

The Minimum Deviation Trap

Proof by Contradiction

Similar Questions

The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Which of the following graphs is the correct one, if $D_{m}$ is the angle of minimum deviation ?

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A prism of angle $A = 1^{\circ}$ has a refractive index $μ = 1.5$ . A good estimate for the minimum angle of deviation (in degree) is close to $N /10$ . Value of $N$ is ……… .

Monochromatic light is incident on a glass prism of angle $A$ . If the refractive index of the material of the prism is $μ$ , a ray incident at an angle $θ$ , on the face $A B$ would get transmitted through the face $A C$ of the prism provided

For an isosceles prism of angle $A$ and refractive index $μ$ , it is found that the angle of minimum deviation $δ_{m} = A$ . Which of the following options is/are correct?

The refractive index of glass is $1.520$ for red light and $1.525$ for blue light. Let $D_{1}$ and $D_{2}$ be angles of minimum deviations for red and blue light respectively in a prism of this glass. Then,

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Visualized Solution

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

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

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

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

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

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

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

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

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