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! :)
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.
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.
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.
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.
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!
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
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
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.
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.
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?
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.
Take a look at the videos in this playlist: PHYSICS 55.1 LENSES AND MIRRORS UNDERSTOOD th-cam.com/play/PLX2gX-ftPVXWwrzHuYCwr0jzl-OpppI6y.html The describe what happens when we change the object distance.
EJ Lee
Thank you for letting me know and congratulations on your excellent performance.
This series has helped me understand the convention and appreciate the topic a lot more, thank you so much.
Great. Glad you found our videos. 🙂
This is a number one Lecture set for Physics of Optics. Thank you very much teacher.
Thank you. Glad you liked it. 🙂
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! :)
Your video really useful for me, thank you❤
very good teacher indeed..thank you sir
Your videos are the best! Very relevant to college lectures, thank you!
Thank you so much 💕 for this
You're so welcome!
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.
Even though the second image used a virtual image as its object, image 2 is still considered a real image?
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.
Thanks, I got it now. Just so many semantics with optics!
THANKYOU SO MUCH
You are welcome.
Is it not possible to draw the ray diagram showing the formation of the second image?
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.
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.
Mhlengeni Miya
Good luck!
Michel van Biezen I will be waiting for this videos on Optics subjects!!! Thank You for all the help you are giving... : )
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!
Since the I1 is virtual, why isn't s2 negative?
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
@@MichelvanBiezen Okay that makes more sense. Thank you!
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
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.
Thank you sir!
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.
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.
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?
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.
Michel van Biezen Thanks Michel, I'll take a look at tweaking my model.
Can you help me solve a similar problem but with the object placed at the focal length of the first lens .?
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
@@MichelvanBiezen Thank you Sir !
since they are converging lenses, i am wondering why you said when the ray passes through the lense, it diverges...
+Z Rita
Yes, I meant to say: "change direction"
so converging lense converges, right?
+Z Rita
The rays passing through a converging lens converge.
Thank u for the quick response and ur video!
What would happen if we put a object further from the focal lenght?
Take a look at the videos in this playlist: PHYSICS 55.1 LENSES AND MIRRORS UNDERSTOOD th-cam.com/play/PLX2gX-ftPVXWwrzHuYCwr0jzl-OpppI6y.html The describe what happens when we change the object distance.
thank you!
Wow this is legit
Glad you liked it.