Now this one is a little bit more complicated.

現在這個問題有點複雜。

Notice that we have a beam of light traveling through air coming on a boundary between air and glass.

請注意，我們有一束光穿過空氣，來到空氣和玻璃之間的邊界上。

Here's a slab of glass that's 10 centimeters thick.

這是一塊 10 釐米厚的玻璃板。

As it enters the glass, of course glass has a different index of refraction equal to 1.6 as opposed to equal to 1 for air.

當然，玻璃的折射率為 1.6，而空氣的折射率為 1。

You can see that the light will bend, it will refract towards the normal so that the angle of refraction, theta sub 2, is smaller than the angle of incidence, theta sub 1.

你可以看到，光線會彎曲，向法線折射，這樣折射角（θ sub 2）就小於入射角（θ sub 1）。

Then it travels across the slab, then it reaches the second boundary where you go back from glass into air.

然後，它穿過石板，到達第二個邊界，從玻璃回到空氣中。

And then it turns out since now the beam is traveling from a region where the index of refraction is larger to a region where the index of refraction is smaller, then the light will now bend away from the normal.

然後事實證明，由於現在光束從折射率較大的區域射向折射率較小的區域，是以現在光線會偏離法線彎曲。

And it turns out since this is air and this is air on both sides of the slab, that the direction of the beam after it leaves the slab will be exactly the same as the direction of the beam as it enters the slab.

事實證明，由於這是空氣，而且是板兩側的空氣，光束離開板後的方向與光束進入板時的方向完全相同。

The only difference is that because of the refraction, the difference, the change in direction here, there will be an offset if the beam had traveled straight across to where the beam is now.

唯一不同的是，由於這裡的折射、差異和方向的改變，如果光束直接穿過現在的位置，就會產生偏移。

There will be a difference, a distance, and your job here is to find out what that distance is, how far has that beam been offset by traveling through a 10 centimeter thick slab of glass with an index of refraction of 1.6.

這裡會有一個差值，一個距離，你的任務就是找出這個距離是多少，光束穿過 10 釐米厚、折射率為 1.6 的玻璃板後偏移了多遠。

So how do you do that?

那麼，如何做到這一點呢？

Well the first thing we need to do is find out what theta sub 2 is.

那麼，我們首先要做的就是找出 Theta sub 2 是什麼。

And so therefore we're going to use Snell's Law.

是以，我們將使用斯涅耳定律。

So n1 sine of theta 1 is equal to n2 sine of theta sub 2.

是以，θ 1 的正弦值 n1 等於θ 子 2 的正弦值 n2。

Solving this for theta sub 2, we're going to switch the equation around so we have n2 sine of theta 2 equals n1 sine of theta 1.

在求解 theta 子 2 時，我們將等式調換一下，這樣，θ 2 的正弦值 n2 等於θ 1 的正弦值 n1。

Dividing both sides by n2, we get sine of theta 2 equals n1 over n2 times sine of theta 1.

將兩邊都除以 n2，就得到正弦θ 2 等於 n1 乘以 n2 再乘以正弦θ 1。

And finally, theta sub 2 is equal to the arc sine, the inverse sine, of n1 over n2 times sine of theta sub 1.

最後，θ sub 2 等於 n1 乘以 n2 的弧正弦，即反正弦乘以θ sub 1 的正弦。

Plugging in the numbers to find out what that is equal to, this is equal to the arc sine of n1, which is 1, divided by n2, which is 1.6, times the sine of theta 1, which was given as 30 degrees.

輸入數字以找出等於多少，這等於 n1（即 1）的弧正弦除以 n2（即 1.6），再乘以θ 1（即 30 度）的正弦。

And so this angle is equal to, here's my calculator, so the sine of 30 is 0.5 divided by 1.6.

所以這個角度等於，這是我的計算器，30 的正弦值是 0.5 除以 1.6。

And then take the arc sine of that, inverse sine, and we get 18.21 degrees.

然後取其弧形正弦，即反正弦，得到 18.21 度。

So theta sub 2 equals 18.21 degrees.

是以，θ 子 2 等於 18.21 度。

I added an extra significant figure so I don't have runoff error at the end.

我加了一個額外的有效數字，這樣就不會在最後出現徑流錯誤。

Now, the ray travels across the glass at a slight angle, so this now will become the angle of incidence of the second boundary, so let's call that theta sub 3.

現在，射線以一個微小的角度穿過玻璃，是以這個角度將成為第二條邊界的入射角，我們稱其為 theta sub 3。

Now what is theta sub 3 equal to?

現在，θ sub 3 等於多少？

Well notice that we have two parallel lines here that are bisected by this ray of light right here, and so these two angles here are considered therefore to be what we call alternate, that's what it is, alternate interior angles.

請注意，這裡有兩條平行線被這條光線一分為二，是以這兩個角被認為是我們所說的交替角，就是交替內角。

I just lost it.

我只是失去了它。

So they're called alternate interior angles and, knowing geometry, those two angles therefore must be equal to each other.

是以，它們被稱為交替內角，根據幾何知識，這兩個角必須相等。

So theta sub 3 is therefore equal to theta sub 2, which therefore is equal to 18.21 degrees.

是以，θ sub 3 等於θ sub 2，而θ sub 2 等於 18.21 度。

Then we can see that the beam will then refract outward, and this will then be theta sub 4, and as we said before, theta sub 4 is expected to be the same as theta sub 1, but just so you can see that that's indeed the case, let's go ahead and calculate that.

然後，我們可以看到光束會向外折射，這將是 theta sub 4，正如我們之前所說，theta sub 4 預計與 theta sub 1 相同，但為了讓大家看到事實確實如此，讓我們繼續計算一下。

So here we're going to say that N3 sine of theta sub 3 is equal to N4 sine of theta sub 4.

是以，我們可以說 N3 正弦的θ 次 3 等於 N4 正弦的θ 次 4。

And so to help us with the index of refractions, notice that we're now going to be working on this boundary right here, so we're going to call this N sub 3, we're going to call this N sub 4, and N sub 3 is equal to 1.6, and N sub 4 is equal to 1, just so we keep things straight.

為了幫助我們計算折射率，請注意，我們現在要研究的是這裡的邊界，所以我們要把它叫做 N 子 3，把它叫做 N 子 4，N 子 3 等於 1.6，N 子 4 等於 1，這樣我們就能把事情弄清楚了。

Coming back over here, we're going to solve for sine of theta sub 4, so we have N4 sine theta sub 4 equals N3 sine of theta sub 3, so we have the sine of theta sub 4 equals N3 over N4 times sine of theta sub 3, and finally, theta sub 4 is equal to the arc sine of N3 over N4 times sine of theta sub 3.

回到這裡，我們要求解θ sub 4 的正弦，是以θ sub 4 的 N4 正弦等於θ sub 3 的 N3 正弦，所以θ sub 4 的正弦等於 N3 乘以 N4 再乘以θ sub 3 的正弦，最後，θ sub 4 等於 N3 乘以 N4 再乘以θ sub 3 的弧形正弦。

There we go.

好了

And plug in the numbers, that's equal to the arc sine of N3 is now 1.6, N4 is 1, times the sine of 18.21 degrees.

輸入數字，等於 N3 的弧正弦現在是 1.6，N4 是 1，乘以 18.21 度的正弦。

Let's see what that is equal to, of course we expect that to be 30 degrees, let's find out.

讓我們來看看這等於多少度，當然我們希望是 30 度，讓我們來看看。

So we take the sine of that, we multiply the times 1.6, and we take the arc sine of that, and sure enough, that's equal to 30 degrees, that's theta sub 4.

是以，我們取其正弦，乘以 1.6，再取其弧形正弦，果然等於 30 度，這就是 theta sub 4。

That's the exiting angle, the angle of refraction at the second boundary.

這就是出射角，也就是第二邊界的折射角。

Alright, now that we know that, we now need to figure out what D is equal to, so before we start with D, let's figure out what this distance is equal to, right here.

好了，既然我們已經知道了這一點，現在就需要計算出 D 等於多少，所以在開始計算 D 之前，我們先來計算一下這個距離等於多少，就在這裡。

So let's call this distance right here X, and to find what X is equal to, let's draw this line all the way down here, and let's look at this big triangle first, so let's call this X sub, now let's call it X, and then this distance right here, let's call that Y, how about Y?

所以我們把這裡的距離叫做 X，為了找到 X 等於多少，我們把這條線一直畫到這裡，先看這個大三角形，所以我們把它叫做 X 子，現在我們把它叫做 X，然後這裡的距離，我們把它叫做 Y，Y 怎麼樣？

Just any old variable will do, we'll just call it Y.

任何一個變量都可以，我們就叫它 Y。

So this distance right here is considered X, this distance right there is considered Y.

是以，這裡的距離被認為是 X，那裡的距離被認為是 Y。

So I think we can find Y first because we know that this distance here is 10 centimeters, we know that the angle right there between this line and that line is 30 degrees, so let's draw this triangle here on the side, so we have this triangle right here, this line right there, this triangle right here, we know that this here is 30 degrees, we know that this here is 10 centimeters, and this distance right here is equal to Y.

所以我想我們可以先找到 Y，因為我們知道這裡的距離是 10 釐米，我們知道這條線和那條線之間的夾角是 30 度，所以讓我們在邊上畫出這個三角形，這樣我們就有了這個三角形、這條線、這個三角形，我們知道這裡是 30 度，我們知道這裡是 10 釐米，這裡的距離等於 Y。

Alright, so how do we find Y?

好吧，那我們怎麼找到 Y 呢？

Well we have the opposite side, we have the adjacent side, we have the angle, so we can say that the tangent of 30 degrees is equal to the opposite side, which is Y, over the adjacent side, which is 10 centimeters, so Y is equal to 10 centimeters times the tangent of 30 degrees, and so therefore Y is equal to 30, take the tangent of that, and times 10 equals, and we have 5.77 centimeters.

我們有對邊，我們有鄰邊，我們有角度，所以我們可以說 30 度的切線等於對邊，也就是 Y，大於鄰邊，也就是 10 釐米，所以 Y 等於 10 釐米乘以 30 度的切線，所以 Y 等於 30，取其切線，乘以 10 等於，我們得到 5.77 釐米。

Alright, so if we know what that is, then we can go ahead and figure out what this distance is, and let's call this distance Z.

好吧，如果我們知道了這是什麼，那麼我們就可以繼續計算出這個距離是多少，讓我們把這個距離稱為 Z。

I'm going to start running out of variables pretty soon, because then you realize that finally we can say that X is equal to Y minus Z.

我很快就要用完變量了，因為你會意識到，我們終於可以說 X 等於 Y 減 Z 了。

So how do we find Z?

那麼，我們如何找到 Z 呢？

Well Z can be found by using a different triangle.

那麼 Z 可以通過使用不同的三角形來找到。

Right here, we can use this triangle.

在這裡，我們可以使用這個三角形。

This is now theta sub 2, and theta sub 2 we found to be 18.21 degrees.

現在是 Theta sub 2，我們發現 Theta sub 2 為 18.21 度。

We know that this vertical side is still 10 centimeters, and this right here now becomes the side Z that we're trying to find out.

我們知道這個垂直面仍然是 10 釐米，而這裡的這個現在就變成了我們要找出的 Z 邊。

Alright, so let's find Z in this case, we know that the tangent of theta sub 2 is equal to the opposite side, which is Z, divided by the adjacent side, which is 10 centimeters, or Z is equal to 10 centimeters times the tangent of theta sub 2, which is 18.21 degrees.

我們知道，θ sub 2 的切線等於對邊（即 Z）除以鄰邊（即 10 釐米），或者說，Z 等於 10 釐米乘以θ sub 2 的切線，即 18.21 度。

So Z is equal to, so we have 18.21, take the tangent of that, and multiply that times 10 centimeters, and we have Z to be 3.29 centimeters.

是以，Z 等於 18.21，取其正切值乘以 10 釐米，Z 就是 3.29 釐米。

Okay, so now we found Y, we found Z, now we can find X, because X is simply going to be Y minus Z, so let me write that here, X is equal to Y minus Z, and Y is 5.77 centimeters, and Z is 3.29 centimeters, so 5.77 minus 3.29 equals 2.48 centimeters.

好了，現在我們找到了 Y，找到了 Z，現在我們可以找到 X，因為 X 就是 Y 減 Z，所以讓我寫在這裡，X 等於 Y 減 Z，Y 是 5.77 釐米，Z 是 3.29 釐米，所以 5.77 減 3.29 等於 2.48 釐米。

We're almost there, so now we have the value for X, so X now is 2.48 centimeters.

我們就快成功了，現在我們有了 X 的值，所以 X 現在是 2.48 釐米。

So how do we find D?

那麼，我們如何找到 D 呢？

Well now we have to make one more triangle.

現在我們得再做一個三角形。

Let me use a slightly different color here, let's use red, so I'm going to draw a line straight across like this, the length of this red line is equal to D, we know the value for this portion right here, which we called X, and by putting that line there I just kind of destroyed my X, so there's my X.

讓我在這裡使用稍微不同的顏色，讓我們使用紅色，所以我要畫一條這樣的直線，這條紅線的長度等於 D，我們知道這裡這部分的值，我們稱之為 X，把這條直線放在這裡，我就破壞了我的 X，所以這就是我的 X。

So we know what X is, we know what D is, and what about this angle right here?

我們知道 X 是什麼，D 是什麼，那麼這個角度呢？

Well this angle has to be equal to this angle, which is equal to 30 degrees, so let me draw this little triangle here on the side, so we have this side here, we have this little line across here, and we have this line across here, and now notice that this line was D, this line was X, and this angle here is 30 degrees, because theta sub 4 is 30 degrees as we found, and this angle must equal this angle.

那麼這個角必須等於這個角，也就是等於 30 度，所以讓我在邊上畫出這個小三角形，所以我們有這邊，我們有這條小線穿過這裡，我們有這條小線穿過這裡，現在注意這條線是 D，這條線是 X，這裡的這個角是 30 度，因為我們發現 theta sub 4 是 30 度，這個角必須等於這個角。

And you say, well why is that so?

你會說，為什麼會這樣呢？

Well notice that this line is perpendicular to this line, and this line is perpendicular to this line, so the angle between those two lines must equal to the angle between those two lines.

那麼請注意，這條直線垂直於這條直線，而這條直線垂直於這條直線，所以這兩條直線之間的夾角一定等於這兩條直線之間的夾角。

And so finding now what D is equal to, notice that X is the hypotenuse, D is the adjacent side, so we could say that D is equal to the hypotenuse X times the cosine of 30 degrees, because D is adjacent to the angle.

注意，X 是斜邊，D 是鄰邊，是以我們可以說，D 等於斜邊 X 乘以 30 度餘弦，因為 D 與角相鄰。

And so finally plugging in what X is equal to right here, which is 2.48 centimeters, so D is equal to 2.48 centimeters times the cosine of 30 degrees, and times 30 cosine equals, and now we found that D is equal to 2.15 centimeters, which is what we're trying to find in the first place.

最後在這裡插入 X 等於 2.48 釐米，所以 D 等於 2.48 釐米乘以 30 度的餘弦，再乘以 30 的餘弦等於，現在我們發現 D 等於 2.15 釐米，這正是我們首先要找到的。

So notice that was not an easy problem.

所以請注意，這不是一個簡單的問題。

Let's recap real quick what we just did.

讓我們快速回顧一下剛才的工作。

We had a light beam traveling through a slab that's 10 centimeters thick.

我們有一束光穿過 10 釐米厚的石板。

It travels across the first boundary, since you're entering a region that has a high index of refraction, it refracts towards the normal, across the second boundary, since you're now traveling from a high index of refraction region to a low index of refraction region, it bends away from the normal, the exiting ray will be parallel to the entering ray, but they'll be offset by some distance D.

穿過第一道邊界時，由於進入的是折射率較高的區域，光線會折射向法線；穿過第二道邊界時，由於從折射率較高的區域進入折射率較低的區域，光線會偏離法線，是以射出的光線會與射入的光線平行，但兩者會偏移一定的距離 D。

You have to find out what that is.

你必須弄清楚那是什麼。

Then you have to find the dimensions of three triangles, this one, the big one, the smaller one, and then this one right here in succession, to find what D is equal to.

然後，你必須找出三個三角形的尺寸，這一個、大的一個、小的一個，然後依次是這裡的這個，找出 D 等於多少。

And that's how you use refraction in a case like that.

這就是在這種情況下使用折射的方法。