In this case we discover that there's a very interesting way in which light can be polarized.

在這種情況下，我們發現有一種非常有趣的方法可以使光線偏振。

And that light is associated with something called Brewster's angle.

這種光與一種叫做布魯斯特角的東西有關。

It was discovered that when light shines onto a surface, like a boundary, some of the light will be transmitted, or refracted as we should say, and some of the light will be reflected.

人們發現，當光線照射到像邊界這樣的表面上時，一部分光線會被透射，或者我們應該說是折射，而另一部分光線則會被反射。

So part of the light will be reflected off the surface, part of the light will be transmitted or refracted through the surface.

是以，部分光線會從表面反射，部分光線會通過表面透射或折射。

And of course, if there's a difference in the index of refraction, if this is N1 and N2, and N2 is larger than N1, we realize that if we then draw the normal to the surface, that the angle of incidence, theta sub 1, is going to be larger than the refracted angle, theta sub 2.

當然，如果折射率存在差異，如果是 N1 和 N2，且 N2 大於 N1，我們就會意識到，如果繪製表面的法線，入射角（theta sub 1）將大於折射角（theta sub 2）。

And it turns out that if the light, which is unpolarized as it approaches the boundary, gets reflected in such a way that the angle between the reflected and the refracted angle is equal to 90 degrees, then the light that gets reflected will be completely polarized, just like as if it went through a polarizer.

事實證明，如果光線在接近邊界時是非偏振的，但在反射時，反射角和折射角之間的夾角等於 90 度，那麼被反射的光線就會完全偏振，就像通過偏振器一樣。

Now of course in this example you can see that it's not quite 90 degrees.

當然，在這個例子中，你可以看到並不是完全呈 90 度。

But if we have an example where the light comes in at just the right angle, so that it reflects in such a way that the refracted angle makes a 90 degree angle with the reflected angle, if this is equal to 90 degrees, then the light that is reflected is polarized.

但是，如果我們舉一個例子，光線以恰到好處的角度射入，使其折射角與反射角成 90 度角，如果這個角度等於 90 度，那麼反射的光線就是偏振光。

Quite an interesting phenomenon.

這是一個相當有趣的現象。

And so the angle of the incident light is then considered Brewster's angle.

是以，入射光的角度被認為是布儒斯特角。

What is that angle?

這個角度是什麼？

What is the angle that it needs to be, that the inbound light is at relative to the normal of the surface so that the light that's reflected will be polarized?

入射光線相對於表面法線的角度是多少，這樣反射出來的光線才會偏振？

And of course what we need to know is what n1 and n2 is.

當然，我們需要知道 n1 和 n2 是多少。

And let's assume that this is air, so that n1 is equal to 1 for air.

假設這裡是空氣，那麼空氣的 n1 等於 1。

And if this is, let's say, water, a smooth water surface, then n2 would be equal to 1.33.

比方說，如果這是水，一個光滑的水面，那麼 n2 就等於 1.33。

In this particular case, what would Brewster's angle be equal to?

在這種特殊情況下，布魯斯特的角度等於多少？

Alright, so remember that the condition of the reflected light and the refracted light, that the angle between them should be 90 degrees.

好了，請記住反射光和折射光的條件，它們之間的夾角應該是 90 度。

Now if we continue this normal line right here, and we call this here theta sub 2, and then knowing that this is the incident light, and then this would be the reflected light, theta reflected, which of course has to be equal to theta sub 1, the incident light, which is equal to Brewster's angle.

現在，如果我們在這裡繼續這條法線，並將其稱為 theta sub 2，然後知道這是入射光線，然後這將是反射光線，theta reflected，當然必須等於入射光線 theta sub 1，也就是等於布儒斯特角。

Alright, so since the distance or the angle from there to there is 180 degrees, if you subtract 90 from that, we then see that theta sub 1 and theta sub 2 added together therefore must be equal to 90 degrees.

好吧，既然從那裡到那裡的距離或角度是 180 度，那麼如果從那裡減去 90 度，我們就可以看到 Theta sub 1 和 Theta sub 2 相加一定等於 90 度。

So theta sub 1 plus theta sub 2 must equal 90 degrees.

是以，θ 子 1 加上θ 子 2 必須等於 90 度。

Now if we use Snell's Law, where we can see that n1 sine of theta 1 is equal to n2 sine of theta sub 2, and then realize that theta sub 2 can be written as 90 degrees minus theta sub 1, and then reinsert it into this equation from Snell's Law, we then get n1 sine of theta 1 is equal to n2 times the sine of 90 degrees minus theta sub 1.

現在，如果我們使用斯涅耳定律，我們可以看到 n1 正弦的θ 1 等於 n2 正弦的θ sub 2，然後意識到θ sub 2 可以寫成 90 度減θ sub 1，然後將其重新插入斯涅耳定律的這個等式中，我們就可以得到 n1 正弦的θ 1 等於 n2 乘以 90 度減θ sub 1 的正弦。

And now notice we now have an equation with only theta sub 1 in it.

現在請注意，我們現在有了一個只有 theta sub 1 的方程。

Of course we still have this 90 degrees there which we want to get rid of.

當然，我們還想去掉那裡的 90 度。

So how do we get rid of that 90 degrees minus theta sub 1?

那麼，如何消除 90 度減 Theta 子 1 呢？

Well for that we need to know a trigonometric identity.

為此，我們需要知道三角函數的特性。

We know that the sine of A minus B can be written as the sine of A, cosine of B minus cosine of B, cosine of not B but A, cosine of A, sine of B.

我們知道，A 減 B 的正弦可以寫成 A 的正弦、B 的餘弦減 B 的餘弦、不是 B 而是 A 的餘弦、A 的餘弦、B 的正弦。

Alright, if we then apply that to what we have over here, we can then say that the sine of 90 degrees minus theta sub 1 is equal to the sine of 90 degrees times the cosine of theta sub 1 minus the cosine of 90 degrees times the sine of theta sub 1.

好了，如果我們把它應用到這裡，我們就可以說，90 度的正弦減去 Theta sub 1 等於 90 度的正弦乘以 Theta sub 1 的餘弦再減去 90 度的餘弦乘以 Theta sub 1 的正弦。

And of course the cosine of 90 degrees is equal to zero, so this term disappears and the sine of 90 degrees is equal to 1.

當然，90 度的餘弦等於零，所以這個項消失了，90 度的正弦等於 1。

So we then can say that the sine of 90 degrees minus theta sub 1 is equal to the cosine of theta sub 1.

是以我們可以說，90 度的正弦減去 theta sub 1 等於 theta sub 1 的餘弦。

So in this Snell's Law here we can replace sine of 90 degrees minus theta sub 1 by cosine of theta sub 1.

是以，在斯涅耳定律中，我們可以用θ sub 1 的餘弦值代替 90 度減θ sub 1 的正弦值。

So now we have n1 sine of theta 1 is equal to n2 times the cosine of theta sub 1.

是以，θ 1 的 n1 正弦等於θ sub 1 的餘弦的 n2 倍。

And now again we have a simplified equation where the only variable is theta sub 1, only one variable.

現在我們又有了一個簡化方程，其中唯一的變量是 theta sub 1，只有一個變量。

So we're going to solve that for theta sub 1.

是以，我們將求解θ 子 1。

So we're going to divide both sides by cosine of theta and both sides by n sub 1, which means we now have the sine of theta sub 1 divided by the cosine of theta sub 1 by moving the cosine down here is equal to n sub 2 divided by n sub 1.

是以，我們要將兩邊都除以 theta 的餘弦，再將兩邊都除以 n sub 1，這意味著我們現在有了 theta sub 1 的正弦除以 theta sub 1 的餘弦，將這裡的餘弦向下移動，等於 n sub 2 除以 n sub 1。

And of course the sine divided by the cosine, that's equal to the tangent.

當然，正弦除以餘弦就等於正切。

So now we can say that the tangent of theta sub 1 is equal to n2 over n1, which means that theta sub 1 is equal to the arctangent of n2 over n1.

是以，現在我們可以說，θ sub 1 的正切等於 n2 除以 n1，這意味著θ sub 1 等於 n2 除以 n1 的反正切。

And in this case, n2 was 1.33, n1 was 1.

在這種情況下，n2 是 1.33，n1 是 1。

And so this is equal to the arctangent of 1.33 over 1.

是以，這等於 1.33 除以 1 的餘切值。

And where's my calculator?

我的計算器呢？

Right here.

就在這裡。

Let's find out what that's equal to, 1.33.

讓我們來看看等於多少，1.33。

Take the arctangent of that and we get 53.06 degrees, 53.06 degrees.

取其反正切，得到 53.06 度，53.06 度。

That's the angle required for theta sub 1 to have reflection that will be polarized because at that moment the angle between the reflected light and the refracted light will be equal to 90 degrees and that is then known as Brewster's angle.

這就是 Theta sub 1 發生偏振反射所需的角度，因為此時反射光和折射光之間的夾角等於 90 度，這就是所謂的布儒斯特角。

So in this case, Brewster's angle is equal to 53.06 degrees and when you shine light at that angle relative to the normal on a smooth water surface, the reflected light will be polarized.

是以，在這種情況下，布儒斯特角等於 53.06 度，當你以這個角度相對於法線照射光滑的水面時，反射光將被偏振。

And that's how you do that problem.

這就是解決這個問題的方法。