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Chapter 11: Light and Geometrical Optics

Re: Chapter 11: Light and Geometrical Optics

Postby goldstanda3269 » Sun Aug 21, 2016 9:00 pm

Lavi wrote:Quick question about the focal length- so it is the distance between the centre of a lens or curved mirror and its focus. But what is the focus point?



It is the point where incoming parallel light rays come into focus after going through a thin lens or mirror: https://i.ytimg.com/vi/_LI9eqRk1vY/maxresdefault.jpg

More generally, it is the point at which rays or waves meet after reflection or refraction, or the point from which diverging rays or waves appear to proceed.
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Re: Chapter 11: Light and Geometrical Optics

Postby Lavi » Mon Aug 22, 2016 10:18 pm

Thank you, so is 2F just another point where light rays can meet?
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Re: Chapter 11: Light and Geometrical Optics

Postby goldstanda3269 » Tue Aug 23, 2016 1:06 am

Lavi wrote:Thank you, so is 2F just another point where light rays can meet?


No.

2f = r

In other words, the radius of curvature of the lens or mirror is twice the focal length.

This is a direct solution to:

2/r = 1/f = 1/o + 1/i (see Optics video)

so

2/r = 1/f

meaning

r = 2f
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Re: Chapter 11: Light and Geometrical Optics

Postby Lavi » Tue Aug 23, 2016 1:42 am

Ah ok! I understand, thank you so much!
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Re: Chapter 11: Light and Geometrical Optics

Postby brad.willo2670 » Fri Apr 26, 2019 9:48 am

Hi there

Q16 - Foundational questions

'16) Consider the following ray diagram of light reflected from a concave mirror where the focal point f (i.e. the principal focus) is as indicated. Where along the principal axis is the image located?'

I am failing to understand why the answer is A and the explanation given in the question did not un-muddy the waters for me. I had C as the answer (focal point).

Also what relation, if any, is there between the focal point and the image point.

Thanks,
B
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Re: Chapter 11: Light and Geometrical Optics

Postby goldstanda3269 » Fri Apr 26, 2019 11:46 am

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 ).
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Re: Chapter 11: Light and Geometrical Optics

Postby horbowiec.7527 » Mon May 03, 2021 11:05 am

Hello, I have couple of questions regarding optics chapter:

a) Is convention summary on bottom of page PHY-157 relating to BOTH concave and convex mirrors?

b) Is convention summary at the end of page PHY-161 relating to BOTH concave and convex lenses?

c) Are points 1, 2, 3, 4 and 5 on page PHY-161 relating to BOTH concave and convex lenses?

Also I would like to point out that terminology for Figure III.B.11.1 (PHY-157) includes o (object distance) and i (image distance) which is not included within the Figure III.B.11.1 itself (PHY-156) and makes it a bit confusing to understand what the text is referring to while discussing this figure.

Regards,
Ania
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Re: Chapter 11: Light and Geometrical Optics

Postby goldstanda3269 » Wed May 05, 2021 4:11 am

horbowiec.7527 wrote:Hello, I have couple of questions regarding optics chapter:

a) Is convention summary on bottom of page PHY-157 relating to BOTH concave and convex mirrors?



Yes, definitely. However, we will see that the same conventions are true for concave/convex mirrors and concave/convex lenses.


horbowiec.7527 wrote:b) Is convention summary at the end of page PHY-161 relating to BOTH concave and convex lenses?


Yes, and for mirrors as previously mentioned.



horbowiec.7527 wrote:
c) Are points 1, 2, 3, 4 and 5 on page PHY-161 relating to BOTH concave and convex lenses?


Yes, and mirrors.


Yes, you are correct that the images do not include i and o, and this will be updated in the next edition.

However, I think that you would find it very helpful to watch the Optics videos that you have access to. If you have not found them, let us know.
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Re: Chapter 11: Light and Geometrical Optics

Postby horbowiec.7527 » Wed May 05, 2021 9:49 am

Thank you.
I always do book chapter first than watch videos so yes I did find the videos and it was more clear!

Cheers,
Ania
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Re: Chapter 11: Light and Geometrical Optics

Postby goldstanda3269 » Thu May 06, 2021 2:02 pm

horbowiec.7527 wrote:Thank you.
I always do book chapter first than watch videos so yes I did find the videos and it was more clear!

Cheers,
Ania


You are making very good progress!

Just in case you have not read these tips yet (either from the following link, or from your Masters Series Book 1, RHSS-07 to RHSS-12), here are some suggestions regarding the quality of your personal notes that you are creating as you study ... https://www.gamsat-prep.com/gamsat-stud ... nal-notes/
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Re: Chapter 11: Light and Geometrical Optics

Postby emstevens46670 » Wed Aug 25, 2021 7:48 pm

Hi there,

1. In relation to Question 7, why isn't the distance to the left of the mirror? The answer says that it is to the right of the mirror.

2. In relation to Question 10, isn't the worked solutions talking about concave mirror not the convex mirror?

I was under the impression that the the concave mirror is always virtual, upright (= erect) and reduced (smaller than the object) and not the convex mirror.

Thank you so much,

Kind regards and best wishes,

Em Stevens
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Re: Chapter 11: Light and Geometrical Optics

Postby goldstanda3269 » Thu Aug 26, 2021 10:59 am

emstevens46670 wrote:
1. In relation to Question 7, why isn't the distance to the left of the mirror? The answer says that it is to the right of the mirror.


The worked solution is extremely detailed. We also have a diagram in the worked solution showing why the image is on the right of that mirror.

Also, note that in your book in the section about reflection (PHY 11.3), page PHY-157 in the Masters Series, it says about an image for a convex mirror: REV which stands for Reduced, Erect and Virtual. Because it must be virtual, it must be on the OPPOSITE side of the mirror from the real object.
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Re: Chapter 11: Light and Geometrical Optics

Postby goldstanda3269 » Thu Aug 26, 2021 11:05 am

emstevens46670 wrote:
2. In relation to Question 10, isn't the worked solutions talking about concave mirror not the convex mirror?

I was under the impression that the the concave mirror is always virtual, upright (= erect) and reduced (smaller than the object) and not the convex mirror.


Your impression is not consistent with the information in the book. Real images can form with concave mirrors and those images can be reduced or enlarged depending on other variables (e.g. f and o).

Side note: this relates back to Question 7, REV for convex mirrors.
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Re: Chapter 11: Light and Geometrical Optics

Postby RosieMcFadyen » Sun Sep 04, 2022 2:14 am

Hi I'm wondering for question 19 how is A wrong? On the explanation it says because the image is inverted but I dont understand how it is inverted while C's image is not inverted. Also how is the Image on the left side of the lens when it is not a mirror?I thought lenses do not reflect the light they only refract
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Re: Chapter 11: Light and Geometrical Optics

Postby goldstanda3269 » Wed Sep 07, 2022 5:22 pm

RosieMcFadyen wrote:Hi I'm wondering for question 19 how is A wrong? On the explanation it says because the image is inverted but I dont understand how it is inverted while C's image is not inverted.


Of course, inverted means 'upside down'. A common symbol to represent this change in optics is to represent the object as an arrow pointing upward (this is also used in multiple ACER GAMSAT practice questions). In so doing, if the image is an arrow pointing in the opposite direction (e.g. pointing down), then it is inverted as is the case for answer choice A which suffers from 2 problems: it should not be inverted AND inverted images must be below the axis (the dark horizontal line).


RosieMcFadyen wrote:I thought lenses do not reflect the light they only refract


You are essentially correct. Please keep in mind the definition of a real image: Light actually passes through the image point. If it is a mirror (which reflects) then a real image MUST be on the same side of the mirror as the real object since light does not pass through the mirror. Likewise, if it is a lens (which refracts), a real image MUST be on the opposite side of the lens from the real object since light passes through the lens.

And so, if an image is behind a mirror, it is virtual. If an image is on the same side of a lens as the real object, then it is virtual.


Regarding the optics rules used in question 19, many of these rules for a diverging lens were needed to solve ACER's GAMSAT Practice Test 2, Section 3, Question 43.
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