Now let's do a quick mirror problem. How close should you stand to a converging funhouse mirror to make your image 75% bigger than you? Now, 75% bigger means that our magnification is equal to 1.75. Since negative image over objects equals magnification, this also equals 1.75.

Now 1.75 is the same as 7/4 or one and three-quarters. We can then rewrite this to say that i equals negative 7/4 of the object. And now let's take our thin spherical mirror equation 1over object plus 1 over image equals 1 over focal length. We'll plug this in. 1 over object plus 1 over negative 7/4th object equals 1 over focal length.

This will flip into the top. 1 over object plus negative 4 over 7 object equals 1 over focal length. And now we've gotta add these. We have to multiply this and this by seven, top and bottom. We get seven minus four over seven objects equals one over focal length. Seven minus four is three, so three over seven objects equals one over focal length.

We rearrange this, we can say that the focal length equals 7/3rds of the object or the object should be at a distance of 3/7ths of the focal length. So if we have a thin, spherical mirror with a focal length, f, we would put our object at 3/7ths of that focal length. Now let's do a Snell's law problem.

Snell's law says n1 sine theta 1 equals n2 sin theta 2. Now we have all of these numbers except for n2, which we're gonna solve for. So let's plug them in. 1.5 is the same thing as 3 divided by 2, or three-halves. So 3 over 2 times the sine of theta1, which is 45 degrees, equals n2 times the sine of 30 degrees.

Now remember sine of 45, sine or 30, these are values we want to either memorize or be able to quickly arrive at using the triangles, the unicircle. The sine of 45 is the square root of 2 over 2. Sine of 30 is 1 over 2, or one-half. So we've got 3 halves times the square root of 2 over 2 equals n2 times one-half. We'll multiply both sides by 2.

And we can see that n2 equals three-halves root 2. If we were to draw this, N1 is 1.5, n2 is essentially 1.5 times root 2. This number's larger. Does that make sense for n2 to be greater than n1? Well, as our light ray comes in here it bends toward the normal, which indicates the refractive index is larger.

So our n2 should be larger than our n1. Now let's do a lens problem. How far from the lens would you place an object to generate an image of the same size? Now, if it doesn't matter if the image is upside down or right side up, then we can just say that for a magnification of negative i over o, if image equals object, if it's going to be upside down the image would have a negative magnification, negative one.

If it's right side up it would just be positive 1. We'll just use image equals object and then we plug this into our equation, 1 over object plus 1 over image equals 1 over focal length. The thin spherical lens equation is extremely useful, always remember it. Now we're going to say that the object and the image links are equal, so 1 over o plus 1 over o equals 1 over focal length.

We add these two together we get, 2 over object equals, 1 over focal length or 2 times the focal length equals the object. So if we have a lens with the focus here, if we place the object in a distance 2F, It will have an equal size. It'll just be an opposite direction. We can plug this back in and see.

If 1 over the object is equal to 1 over 2F plus 1 over the image equals 1 over focal length, then we can subtract that, one over the image equals one over f minus one over 2f multiply both sides by 2 here, we get 1 over the image equals 2 minus 1 over 2f or 1 over 2f. So we can just rewrite this image equals 2f, the same.

Read full transcriptNow 1.75 is the same as 7/4 or one and three-quarters. We can then rewrite this to say that i equals negative 7/4 of the object. And now let's take our thin spherical mirror equation 1over object plus 1 over image equals 1 over focal length. We'll plug this in. 1 over object plus 1 over negative 7/4th object equals 1 over focal length.

This will flip into the top. 1 over object plus negative 4 over 7 object equals 1 over focal length. And now we've gotta add these. We have to multiply this and this by seven, top and bottom. We get seven minus four over seven objects equals one over focal length. Seven minus four is three, so three over seven objects equals one over focal length.

We rearrange this, we can say that the focal length equals 7/3rds of the object or the object should be at a distance of 3/7ths of the focal length. So if we have a thin, spherical mirror with a focal length, f, we would put our object at 3/7ths of that focal length. Now let's do a Snell's law problem.

Snell's law says n1 sine theta 1 equals n2 sin theta 2. Now we have all of these numbers except for n2, which we're gonna solve for. So let's plug them in. 1.5 is the same thing as 3 divided by 2, or three-halves. So 3 over 2 times the sine of theta1, which is 45 degrees, equals n2 times the sine of 30 degrees.

Now remember sine of 45, sine or 30, these are values we want to either memorize or be able to quickly arrive at using the triangles, the unicircle. The sine of 45 is the square root of 2 over 2. Sine of 30 is 1 over 2, or one-half. So we've got 3 halves times the square root of 2 over 2 equals n2 times one-half. We'll multiply both sides by 2.

And we can see that n2 equals three-halves root 2. If we were to draw this, N1 is 1.5, n2 is essentially 1.5 times root 2. This number's larger. Does that make sense for n2 to be greater than n1? Well, as our light ray comes in here it bends toward the normal, which indicates the refractive index is larger.

So our n2 should be larger than our n1. Now let's do a lens problem. How far from the lens would you place an object to generate an image of the same size? Now, if it doesn't matter if the image is upside down or right side up, then we can just say that for a magnification of negative i over o, if image equals object, if it's going to be upside down the image would have a negative magnification, negative one.

If it's right side up it would just be positive 1. We'll just use image equals object and then we plug this into our equation, 1 over object plus 1 over image equals 1 over focal length. The thin spherical lens equation is extremely useful, always remember it. Now we're going to say that the object and the image links are equal, so 1 over o plus 1 over o equals 1 over focal length.

We add these two together we get, 2 over object equals, 1 over focal length or 2 times the focal length equals the object. So if we have a lens with the focus here, if we place the object in a distance 2F, It will have an equal size. It'll just be an opposite direction. We can plug this back in and see.

If 1 over the object is equal to 1 over 2F plus 1 over the image equals 1 over focal length, then we can subtract that, one over the image equals one over f minus one over 2f multiply both sides by 2 here, we get 1 over the image equals 2 minus 1 over 2f or 1 over 2f. So we can just rewrite this image equals 2f, the same.