brad.willo2670 wrote:Also what relation, if any, is there between the focal point and the image point.

The relation is the most important equation in all of optics (i = image dist; o = object dist):

1/f = 1/i + 1/o

Sometimes ACER gives that equation in exams, but sometimes they do not. In our Optics video, it is written this way:

2/r = 1/f = 1/i + 1/o

Which reminds you that the radius of curvature r is twice the focal length f: r = 2f.

Question 16 does not need to be done mathematically, but just for fun:

We are given that f is at C. Let's choose a random number for f as 2. Now B is clearly more than 2 but less than 4, so let's choose 3. Now let's find the image (actually, this is a common, basic GAMSAT calculation which you should consider doing yourself before looking at the answer):

1/2 = 1/i + 1/3

3/6 = 1/i + 2/6

1/6 = 1/i

i = 6 and now you see that the answer must be to the left of B making A the only option.

Now let's examine the problem the faster way: For an overview, please look carefully at the image at A in the explanation and notice the arrow is larger and upside down.

Also notice that parallel light rays (meaning, parallel to the central horizontal line called the 'central axis') are related to the focal point f (which is a point on the central axis measured by the equation above and clearly 1/2 the distance to the radius of curvature of the mirror or lens).

Notice that the light rays START at the top of the arrow, well, wherever those light rays converge (come together) that's where you will see the image of the top of the arrow. That's why, without even calculating, you can know that the image forms BELOW A.

In the Purple Booklet by ACER (GAMSAT Practice Test 2), Questions 40 – 43, all of the concepts above are explored (I'm sure you know, we put the worked solution on Youtube:

https://youtu.be/MYu9EC5oeGI ).