Question Video: Calculating Indices of Refraction from the Dispersive Power of a Prism Physics

Video Transcript

A prism has a dispersive power of 0.076. White light is dispersed by the prism. For the longest wavelength of light passing through it, the prism has a refractive index of 1.37. What is the refractive index of the prism for the shortest wavelength of light passing through it? Answer to two decimal places.

So for this question, we’re told the dispersive power of a prism, and we’re also told the refractive index for the longest wavelength of light passing through it. Given this information, we’re asked to calculate the refractive index of the prism for the shortest wavelength of light passing through it. Let’s say that this triangle represents the prism in the question and this thick arrow represents white light entering the prism. Then we can recall that a prism with different refractive indices for different wavelengths of light will disperse or spread out the different colors that make up the white light passing through it. The dispersive power of the prism is a number that measures how much the prism spreads out these different colors of light. The more the prism spreads out the light, the larger this number is.

We usually denote the dispersive power of a prism with the symbol 𝜔 𝛼. And for this question, we’re told that the dispersive power of our prism is 0.076. So we can write down that 𝜔 𝛼 equals 0.076 for this prism. We’re also told the refractive index for the longest wavelength of light passing through the prism. And since we know that red light has the longest wavelength out of all the colors of visible light, we know that this is the refractive index for red light. So we’re told that the refractive index for red light is equal to 1.37. And we usually denote this refractive index by the symbol 𝑛 subscript min because we know that since red light is deviated the least by the prism, it must have the minimum refractive index out of all the colors of light being refracted by the prism.

On the other hand, we know that blue light, which has the shortest wavelength out of all the colors that make up white light, is deviated the most by the prism. So we know that it must have the maximum refractive index. Because of this, we call the refractive index for blue light 𝑛 max. And this is precisely the quantity we want to calculate for this question. So let’s clear some space on the screen and see how we can calculate this refractive index. So we’ve got the quantities we’re told in the question: the dispersive power of the prism 𝜔 𝛼 is equal to 0.076, and the refractive index for the longest wavelength of light, which we call 𝑛 min, is equal to 1.37. And we want to find the refractive index for the shortest wavelength of light 𝑛 max.

So let’s begin by recalling the equation that links these three quantities together. This equation reads 𝜔 𝛼, the dispersive power of the prism, equals 𝑛 max minus 𝑛 min divided by 𝑛 max plus 𝑛 min divided by two minus one. Since the quantity we want to calculate is 𝑛 max, our goal will be to rearrange this equation so that 𝑛 max is the subject of the equation. However, since 𝑛 max appears twice in the right-hand side of this equation, both in the top and the bottom of the fraction, there’ll be quite a few steps to rearranging this equation to get 𝑛 max to be the subject. So, to simplify this a little bit, let’s start by substituting in the values we do know.

We’re told the dispersive power of the prism 𝜔 𝛼 is equal to 0.076. And since the left-hand side of our equation is just 𝜔 𝛼, we can write that the left-hand side of the equation is just equal to 0.076. We also know that 𝑛 min is equal to 1.37. So we can replace the two 𝑛 mins in the right-hand side of our equation with this value. Doing this gives us an equation that looks like this. And now the only unknown variable we have is 𝑛 max, which is exactly what we want to find. Let’s now try to simplify the denominator of the right-hand side of our equation. And we can begin by splitting up the fraction that appears in the denominator into two. That is, we can write the fraction 𝑛 max plus 1.37 divided by two as two quantities: 𝑛 max divided by two plus 1.37 divided by two.

Next, we can calculate 1.37 divided by two and replace it with its numerical value, which is 0.685. The last simplification we can make before we start rearranging this equation is to combine the two numbers in the denominator, 0.685 minus one, which we can calculate to be negative 0.315. So we now have an equation where 𝑛 max is the only unknown variable left, and this equation reads 0.076 equals 𝑛 max minus 3.7 divided by 𝑛 max over two minus 0.315. And we’re now ready to rearrange this equation to make 𝑛 max the subject.

The first step of this rearranging is to multiply both sides of the equation by everything that appears in the denominator of the right-hand side. So we multiply both sides of the equation by 𝑛 max divided by two minus 0.315. The reason we do this is to simplify the right-hand side of the equation. Since the term 𝑛 max divided by two minus 0.315 appears in both the numerator and the denominator of the right-hand side, this expression cancels out. And we’re left with a right-hand side that simply reads 𝑛 max minus 1.37. So our simplified equation now reads 𝑛 max divided by two minus 0.315 all multiplied by 0.076 equals 𝑛 max minus 1.37.

Our next step is to start simplifying the left-hand side of our equation by multiplying out our brackets. And we do this by multiplying both terms in our brackets by 0.076. This gives us two terms on the left-hand side, 0.076 times 𝑛 max divided by two minus 0.076 times 0.315. There are now some numbers we can combine on the left-hand side. So let’s start by combining the 0.076 divided by two we have in the first term. This combines to 0.038, so our first term reads 0.038 times 𝑛 max. And we can also calculate the multiplication 0.076 times 0.315, which we find to be equal to 0.02394.

Next, we want to get both terms involving 𝑛 max on the same side of our equation and both the terms that are just numbers on the other side of our equation. We can start by adding 1.37 to both sides of the equation. We do this because we now have minus 1.37 plus 1.37 on the right-hand side of our equation. So this cancels to zero. And on the left-hand side, we can combine the two numerical terms minus 0.02394 plus 1.37, which together give us a numerical term that reads plus 1.34606. Let’s move this equation up our screen a little bit and now work to get both the terms involving 𝑛 max on the right-hand side of the equation.

We do this by subtracting the term 0.038 times 𝑛 max from both sides of the equation. This gives us 0.038𝑛 max minus 0.038𝑛 max on the left-hand side, so these two terms cancel to zero. And on the right-hand side, we have both terms involving 𝑛 max, and this is the only place that 𝑛 max appears in our equation. And our equation now reads 1.34606 equals 𝑛 max minus 0.038 times 𝑛 max. Since the right-hand side only involves terms that include 𝑛 max, let’s factorize out 𝑛 max from these two terms. This is done by writing down 𝑛 max and then multiplying it by all the quantities that appear in front of 𝑛 max in the line above. So we have a one from this term that’s just 𝑛 max and minus 0.038 from this term.

This is great because this bracket multiplying 𝑛 max is just numbers. So we can simplify this even further if we calculate that one minus 0.038 is equal to 0.962. This almost gives us exactly what we want, which is 𝑛 max all on its own. So to finally get that, let’s divide both sides of this equation by 0.962. This lets us cancel 0.962 from both the top and bottom of the right-hand side, leaving just 𝑛 max. And we finally have an equation for the quantity we want, which is 𝑛 max. So by working out this fraction on our calculator, we can write down that 𝑛 max is equal to 1.39923 and so on.

This is almost our final answer. But recall that the question asked us to give our answer to two decimal places. If we do this rounding, we can give our answer for the refractive index of the prism for the shortest wavelength of light passing through it as 𝑛 max equals 1.40.