PROBLEM

IRstuff · ‎Aug 28, 2003

On 8/27/2003 6:50:31 PM, DDHARLEY wrote:
>HAS ANYONE BEEN READING THE
>POSTS OR IS THERE A PROBLEM
>WITH YOU RECEIVING THEM

The problem is that you've been posting a large number of problems. This is not a homework service and even though you've shown your work, you asking a lot for us to work out every one of your problems to check your answers.

I would suggest that you check your work against your textbook and only post question about problems that you either don't fully understand or are really not sure about the solution.

TTFN,
Eden

IRstuff · ‎Aug 29, 2003

On 8/28/2003 5:26:22 PM, DDHARLEY wrote:
>THAT IS WHAT I HAVE BEEN DOING
>I AM NOT SENDING YOU EVERY ONE
>OF MY PROBLEMS IF I WERE I
>WOULD BE TYPING FOREVER. THERE
>ARE SOME THINGS I AM CONFUSED
>ON AND I AM USING YOU FOR
>VERIFICATION TO SEE IF I AM ON
>THE RIGHT TRACK I ONLY HAVE
>ONE MORE SECTION ON
>ELECTRICITY AND I WILL BE
>FINISHED
>SORRY IF I BOTHERED YOU I JUST
>THOUGHT YOU WOULD HELP THANKS

We are here to help, but when you post more than 30 problems in a span of two weeks, that's a bit much to ask people who are volunteering their time to spend that much time on one person's questions.

If you are really having that much uncertainty about the homework problems you've posted, then you need to find yourself a tutor or to get additional help from your instructor.

TTFN,
Eden

Wave-disabled · ‎Aug 30, 2003

how come you aways type with cap on?

Wave-disabled · ‎Aug 31, 2003

"I DIDNT THINK IT WOULD TAKE SO MUCH WORK TO CHECK MY PROBLEMS"

Actually, most of the time i avoid the post that is really really long. I don't like to spend so much time reading all of them(I like a short-simple problem, a long one made me wants to go to sleep), perhaps this is ONE of the reasons the helper saying stuff about it after many approach of helping you out with a VERY long post. Additionally, they didn't get pay to do this...

Is there a place for you to post your questions besides this place?

Not that i know of...

Wave-disabled · ‎Sep 01, 2003

I didn't help you any of the problem you posted so there is no point giving me credit. The one you should give credit to is tkhunny, pkalford, eden_mei, and others who i don't know hte name.

DDHARLEY-disabl · ‎Sep 03, 2003

I DONT KNOW HOW TO REACH ALL OF THE PEOPLE WHO
HELPED ME BUT I WOULD LIKE TO THANK THEM I ONLY HAVE 2 MORE TESTS AND THEY ARE ON LIGHT AND ELECTRICITY. I THINK I HAVE THE ELECTRICITY WHIPPED BUT THE PROBLEMS I POSTED ON LIGHT I AM NOT SURE OF. WE WILL SEE EVENTUALLY

IRstuff · ‎Sep 03, 2003

WHICH OF THE FOLLOWING VALUES REPRESENTS AN INDEX OF REFRACTION OF AN ACTUAL MATERIAL

A.0

B.1/4

C.1/2

D.5/4

MY BOOK ONLY HAS A SMALL PARAGRAPH ABOUT THIS

IT SAYS ALL MATERIALS MUST HAVE AN INDEX OF REFRACTION GREATER THAN 1.0

WOULD IT BE D BECAUSE IN FRACTION FORM IT WOULD BE
1.25

IT ISNT EXPLAINED VERY WELL IN THE BOOK

Index of refraction, is usually the ratio of the speed of light in vaccuum to speed of light in the material. Since the speed of light is maximum in a vaccuum, in any other material, its speed must be lower, thus index of refraction is ALWAYS >= 1.

TTFN,
Eden

IRstuff · ‎Sep 03, 2003

13. A VIRTUAL IMAGE PRODUCED BY A LENS IS ALWAYS

A.LARGER THAN THE OBJECT

B.SMALLER THAN THE OBJECT

C.LOCATED IN FRONT OF THE LENS

D.LOCATED IN BACK OF THE LENS

MY BOOK SAYS IT IS SMALLER THAN THE OBJECT AND LOCATED IN THE FRONT OF THE LENS

HOW CAN I HAVE 2 ANSWERS

Virtual images are images that you can only see, but essentially cannot touch.

Virtual images are created by diverging lenses, and the image size cannot be bigger than the actual object, since the magnification always winds up <=1.0.

There must be some sort of typo or misunderstanding about image position, since virtual images are always behind the lens, otherwise, you'd get a real image.

TTFN,
Eden

rfaucher-disabl · ‎Sep 03, 2003

Yes it's true that virtual images are in front of the lens and smaller if its a diverging lens (concave).
But there is a special case of a virtual image forming with a converging lens when the object is real close to the lens. In this special situation, the virtual (right-side-up) image is larger than the object and cannot be projected onto a surface. The situation mentioned in your book with both answers referred specifically to a diverging lens. The question you posted referred to a lens in general, which could include the special case of the converging lens I just mentioned.

There is an analogy with curved mirrors. Shoplifting mirrors in stores show small virtual images. But put your face close (inside the focal point) to a make-up (converging) mirror and your face looks right-side up (virtual) and bigger than life! Backing away from this mirror will result in your reflected image inverting upside down, showing you the real images associated with the other possibilities of reflection.

(Thanks Eden for the cool link. I'll use it in my classroom!)

IRstuff · ‎Sep 03, 2003

You might find this useful, it's a Jave applet of a thin lens:

http://www.phy.ntnu.edu.tw/java/Lens/lens_e.html

TTFN,
Eden

DDHARLEY-disabl · ‎Sep 06, 2003

I AM STILL CONFUSED IN THESE I DONT UNDERSTAND

DDHARLEY-disabl · ‎Sep 08, 2003

SO ARE YOU SAYING THAT THE ANSWER TO THE QUESTION ABOUT THE INDEX OF REFRACTION WOULD BE A WHICH IS O

AND THE QUESTION ABOUT THE VIRTUAL IMAGE WOULD BE ANSWER A WHICH IS LARGER THAN THE OBJECT

IRstuff · ‎Sep 08, 2003

On 9/8/2003 7:02:32 PM, DDHARLEY wrote:
>SO ARE YOU SAYING THAT THE
>ANSWER TO THE QUESTION ABOUT
>THE INDEX OF REFRACTION WOULD
>BE A WHICH IS O
>
>AND THE QUESTION ABOUT THE
>VIRTUAL IMAGE WOULD BE ANSWER
>A WHICH IS LARGER THAN THE
>OBJECT

Index of refraction is ALWAYS greater than 1 for normal materials, because light ALWAYS travels SLOWER in normal materials.

TTFN,
Eden