Question Video: Identifying the Relation between Wavelength and Refractive Index of a Prism | Nagwa Question Video: Identifying the Relation between Wavelength and Refractive Index of a Prism | Nagwa

Question Video: Identifying the Relation between Wavelength and Refractive Index of a Prism Physics

Which of the following formulas describes the relation between the wavelength of incident light and the refractive index of a prism for that specific wavelength? [A] 𝑛 ∝ πœ† [B] 𝑛 ∝ 1/πœ† [C] 𝑛 ∝ 1/βˆšπœ† [D] 𝑛 ∝ 1/πœ†Β²

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Video Transcript

Which of the following formulas describes the relation between the wavelength of incident light and the refractive index of a prism for that specific wavelength? (A) 𝑛 is proportional to πœ†. (B) 𝑛 is proportional to one over πœ†. (C) 𝑛 is proportional to one over the square root of πœ†. Or (D) 𝑛 is proportional to one over πœ† squared.

To think about this, let’s first remember that any material that affects different wavelengths of light differently is called dispersive. So, because a prism refracts different wavelengths of light by different amounts, we know it’s a dispersive material.

To understand this better, let’s look at an example diagram showing how a prism affects different wavelengths of light. Here, we can see two rays of light, one red and one blue, incident on the prism at the same point and at the same angle. We know that the wavelength of red light is greater than that of blue light. We can also see that the red ray is refracted less than the blue ray is. This happens because the prism has a different index of refraction for different wavelengths of light. And this question is asking us to identify how we can represent this relationship mathematically.

To do this, let’s remember that the refractive index, 𝑛, is given by 𝑐, the speed of light in a vacuum, divided by 𝑣, the speed of light in a certain material. We should also remember the wave speed equation, which says that the speed of a wave, such as light, 𝑣, equals 𝑓, the wave’s frequency, times πœ†, its wavelength. So, substituting this expression for 𝑣 into the formula for the index of refraction, we get that 𝑛 equals 𝑐 over 𝑓 times πœ†. And since here we’re only concerned with the relationship between 𝑛 and πœ†, let’s ignore the other two terms, 𝑐 and 𝑓, by setting them equal to one just to hold their place in the equation. Notice that we should also replace the equals sign with this symbol to indicate that we’re no longer strictly equating the right- and left-hand sides of this expression.

Thus, we’ve devised a statement of proportionality between 𝑛 and πœ†. We say that 𝑛 is proportional to one over πœ† or that 𝑛 is inversely proportional to πœ†, meaning that as one quantity increases, the other must decrease. This makes sense because, as we’ve seen in this example diagram, light with a greater wavelength is refracted less by a prism or that light with a smaller wavelength is refracted more. Thus, we should eliminate answer option (A), because we know that 𝑛 and πœ† are not directly proportional. They’re inversely proportional. And because this πœ† term in the denominator isn’t raised to the power of two or one-half, we know that 𝑛 is not inversely proportional to either πœ† squared or the square root of πœ†. So we should eliminate answer options (C) and (D) as well.

Therefore, we know that answer option (B) is correct. The relationship between the wavelength of incident light and the refractive index of a prism for that specific wavelength is given by the expression 𝑛 is inversely proportional to πœ†.

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