Visit ilectureonline.com for more math and science lectures! In this video I will show you how to find the location of the image when the object is placed 50cm away from the 2 converging lenses.
Пікірлер: 47
@guliyevshahriyar Жыл бұрын
This is a number one Lecture set for Physics of Optics. Thank you very much teacher.
@MichelvanBiezen Жыл бұрын
Thank you. Glad you liked it. 🙂
@ClarkieUK Жыл бұрын
This series has helped me understand the convention and appreciate the topic a lot more, thank you so much.
@MichelvanBiezen Жыл бұрын
Great. Glad you found our videos. 🙂
@MichelvanBiezen9 жыл бұрын
EJ Lee Thank you for letting me know and congratulations on your excellent performance.
@ejrabot40329 жыл бұрын
Good day Mr. Michel van Biezen! I want to thank you for all the lectures here especially these lecture videos about optics because I was able to understand my lessons in physics that's why I passed my last physics subject!! THANK YOU! :)
@axlsh1r7y7 жыл бұрын
very good teacher indeed..thank you sir
@Artisticengineer59 жыл бұрын
Your videos are the best! Very relevant to college lectures, thank you!
@leeliantsin5096 жыл бұрын
Your video really useful for me, thank you❤
@aseem17087 ай бұрын
THANKYOU SO MUCH
@MichelvanBiezen7 ай бұрын
You are welcome.
@anibalc.ripollr.96438 жыл бұрын
The issue with the position of I1 was due to the distances from the first lens to its focci are different; the distance to the left is shorter than the distance to the right. I guess.
@hadisquantumchannel77162 жыл бұрын
Fascinating how we use the rays of a virtual object to form a real object again. So somehow the information of rays that don’t exist is passed on.
@MichelvanBiezen2 жыл бұрын
In actuality that is not how it works in the physical world, but doing it that way does offer a good mechanical way to solve the problem. In actuality, the rays go through the first lens and will therefor bend (refract) passing through the firs lens, and those rays then will go through the second lens and bend (refract) some more. In the case of a virtual image placing the second lens there will cause the virtual image to appear in a different place as opposed to only viewing it through the first lens.
@brendakimaiyo55633 жыл бұрын
Thank you so much 💕 for this
@MichelvanBiezen3 жыл бұрын
You're so welcome!
@oscareluciane9 жыл бұрын
Hi Michel, I would like to know what is the relation between the focal distance and the lenses distances when we have 2 or more lenses. I have an exercise where f1>distance-lenses and f2=object-lens2-distance. I can't get to draw the rays/image... This is from Hecht 4th Ed. (exercise 5.33). Help, please? Cheers!
@srividyan41726 жыл бұрын
Sir, if s1 dash is -30 cm, then wouldn't it mean that the image would be on the right side of the lens, instead of the left side, as negative distances mean that it is on the right side? Please help me sir
@MichelvanBiezen6 жыл бұрын
With lenses a positive image distance places the image on the right side of the lens. A negative image distance places the (virtual) image on the left side of the lens (provided the object is on the left side). With mirrors it is the other way around.
@srividyan41726 жыл бұрын
Thank you sir!
@Neekodeos10 жыл бұрын
Even though the second image used a virtual image as its object, image 2 is still considered a real image?
@MichelvanBiezen10 жыл бұрын
Niko, Yes, it does appear strange. This is the method we use to find the final image. But what is really going on is that to the observer on the right side of the 2 lenses, the first lens makes it look like the the "original" object is where the virtual image is formed by the first lens. (To the observer it looks as if there is a "real" object 30 cm in front of the first lens which puts it 40 cm in front of the second lens. Thus the "virtual" image acts like a "real" object 40 cm in front of the second lens which then makes a real image to the right of the second lens. A real image is made when the rays converge on the other side (right side) of the second lens.
@Neekodeos10 жыл бұрын
Thanks, I got it now. Just so many semantics with optics!
@vancouverjohn110 жыл бұрын
I am having some difficulty with a project that requires viewing a cell phone and characters from an approx. distance of 4" from the eye. I can easily place a lens with a diopter of 6 directly in front of the eye and read the characters but the project requires the lens/ lenses to be placed close to the face of the cell phone. Would you recommend a combination using converging and diverging lenses?
@MichelvanBiezen10 жыл бұрын
John, If I get this question correct, you are going to place a cell phone 4 inches in front of your eyes and you want to use a lens (or lens combination) placed directly in front of the cell phone to magnify the object. If that is correct, you will not be able to do so. Use the lens equation (1/f) = (1/s) + (1/s'). Placing a phone at 4 inches will be difficult to see since most people's eyes cannot focus at something that close (10 inches is a comfortable distance for most). Placing a converging lens directly over the cell phone will not give you a magnified image. Holding the converging lens farther away from the cell phone will give you a magnified image farther away. The best result will happen when you put the converging lens directly in front of your eyes, but with the object only 4 inches away, you'll need to use a very powerful lens with a very short focal length. Those are usually small and will not show you enough field of view. So you will be able to see a partial magnified image with a strong converging lens held somewhere between the cell phone and you eye.
@vancouverjohn110 жыл бұрын
Michel van Biezen Thanks Michel, I'll take a look at tweaking my model.
@charityburgess76962 жыл бұрын
Wow this is legit
@MichelvanBiezen2 жыл бұрын
Glad you liked it.
@mhlengenimiya59939 жыл бұрын
Is it not possible to draw the ray diagram showing the formation of the second image?
@MichelvanBiezen9 жыл бұрын
Miya, Using this technique of finding a secondary image it is difficult. The way I do is as follows: find the second image mathematically, and then draw in the ray diagrams from the second lens to the second image. There is a more advanced technique using matrices to follow the rays through each lens and between every lens, but I have not yet made videos on that technique. It is usually taught at the upper division or graduate level.
@mhlengenimiya59939 жыл бұрын
Okay thank you, I was just asking out of curiosity, I have never seen it in the past question papers. Please wish me luck, tomorrow I'm writing my first test on Optics.
@MichelvanBiezen9 жыл бұрын
Mhlengeni Miya Good luck!
@oscareluciane9 жыл бұрын
Michel van Biezen I will be waiting for this videos on Optics subjects!!! Thank You for all the help you are giving... : )
@jfwang9833 жыл бұрын
Since the I1 is virtual, why isn't s2 negative?
@MichelvanBiezen3 жыл бұрын
There is no link between I1 being virtual and what the sign of S2. The sign of S2 only depends on the location of Object 2 (which is I1) relative to lens 2
@jfwang9833 жыл бұрын
@@MichelvanBiezen Okay that makes more sense. Thank you!
@debaleenamajumder31894 жыл бұрын
Can you help me solve a similar problem but with the object placed at the focal length of the first lens .?
@MichelvanBiezen4 жыл бұрын
Take a look at these videos. They will help you understand how to do that: PHYSICS 55.1 LENSES AND MIRRORS UNDERSTOOD and PHYSICS 55.4 MULTIPLE LENSES
@debaleenamajumder31894 жыл бұрын
@@MichelvanBiezen Thank you Sir !
@yxz4588 жыл бұрын
since they are converging lenses, i am wondering why you said when the ray passes through the lense, it diverges...
@MichelvanBiezen8 жыл бұрын
+Z Rita Yes, I meant to say: "change direction"
@yxz4588 жыл бұрын
so converging lense converges, right?
@MichelvanBiezen8 жыл бұрын
+Z Rita The rays passing through a converging lens converge.
@yxz4588 жыл бұрын
Thank u for the quick response and ur video!
@johnnybatafljeska63688 жыл бұрын
What would happen if we put a object further from the focal lenght?
@MichelvanBiezen8 жыл бұрын
Take a look at the videos in this playlist: PHYSICS 55.1 LENSES AND MIRRORS UNDERSTOOD kzbin.info/aero/PLX2gX-ftPVXWwrzHuYCwr0jzl-OpppI6y The describe what happens when we change the object distance.