An object 2cm high is placed 10cm in front of a convex spherical surface of 15cm radius of curvature which separates air from glass of refractive index 1.60. Find the position and size of the image.
An object 2cm high is placed 10cm in front of a convex spherical surface of 15cm radius of curvature which separates air from glass of refractive index 1.60. Find the position and size of the image.
Sorry but my Gauss equation days are long gone and (thankfully!) behind me, but some guru's here will know but may not be willing just give you the answer.
This, for the curious, is what she's asking- Not for the faint of heart*~~~
https://www.khanacademy.org/science/.../convex-lenses
Good luck
* It's actually some of the easiest math that takes place in optics.
Last edited by Uncle Fester; 01-03-2019 at 04:33 PM. Reason: tweek...
Let's think this over!
First, I like the "step vergence" method, because it's intuitive.
Think about light being emanated (via reflection) from this image. It goes out in all directions, like a bubble (ignore the fact that it ricochets off the object, etc. Let's pretend it's the very tip of a pencil we're discussing).
Now, the light rays are very close together as they emanate, like a really prickly porcupine. But as the bubble grows (the "wavefront" is the surface of the bubble), since it's getting bigger, the rays are not so tightly packed. In fact, at about 20 feet bubble radius, it's pretty close to parallel because the wave front is pretty flat.
So, the "packed quiver of porcupine needles or arrows or pickup stix" early phase has high "vergence" (di-vergence, specifically). It's notated by 1/distance [in meters] you are from the object (a "diopter"). So, if the object is 10cm away from the surface of the plastic, the vergence is defined as 1/0.1M or 10D divergent.
Now that light wavefront is affected by the plastic. Some hits the molecules and is reflected, but some goes through the spaces. But the light is SLOWED DOWN by the plastic. Specifically, it is 1.6 times slower (the "index") of the material.
The central ray of the spherical wavefront hits the central point of the spherical lens (be it concave or convex) and it isn't deviated. It slows down upon entry, and speeds up on exit, but isn't deflected off-line. But all the other rays hit at various angles, and are deflected (or "focused", in the case of a sphere), depending on the shape of the lens when light enters and exits.
Cont'd.
If the lens that the light hits would be given to you in focusing power (also known as diopters), then you simply say "Hey, 10D of divergence meets 10D of convergence and the light is made parallel". But they don't do that. They give you the information needed for you to calculate the focusing power yourself.
So we know that focusing power is proportional to how much it slows down light. It began as speed in a vacuum (or, n= 1.0, no effect) (although not really because it's not floating in space), and went to speed n =1.6 (slower), so it is n' - n = 0.6.
We also said that it's proportional to curve of the refracting surface, or, to make matters slightly stranger (same concept, though) inversely proportional to the radius of curvature "r" (in meters) So the formula for refracting power of this surface is n'-n divided by r.
0.6/0.15M = 4D.
Now, light comes in at -10.00D divergent and is converged +4D, so the light leaves the surface still divergent, but only at -6.00D.
Last edited by drk; 01-04-2019 at 02:23 PM.
cont'd.
Now, this problem wants "angular magnification", really. The object looks "x" big in reality, and then the image formed of that object looks "y" big. I ask you, is it going to look larger, or smaller, to your eye, when you look at the object in air, or the object through the lens?
It's a "plus power lens", so that's a clue. From our experience, plus lenses magnify, and minus lenses minify. So it looks bigger to our stupid eyes. And it's a simple ratio.
Size before/size after is proportional to vergence before/vergence after.
So...vergence before is 10. Vergence after is a 4. So it's 10/4 or 2.5 times larger. "2.5X".
So, if it looked 2 cm tall looking right at it, and it looked 2 cm x 2.5 when looking through the lens, or 5 cm.
I think that's right.
Or you could memorize an equation, which is ultimately what you want to do.
https://www.ck12.org/na/lens-formula...ion-x-physics/
Here’s my take on this question. (An object 2cm high is placed 10cm in front of a convex spherical surface of 15cm radius of curvature which separates air from glass of refractive index 1.60. Find the position and size of the image.)
1m = 100cm = 1000mm, so radius of lens curvature = 15cm = 0.15m and index of (air) = n = 1.0
F(power of front surface in diopters) = n’-n/r(m) = 1.60-1.00/0.15m = 0.6/0.15 = +4.00D
Vergence is indeed used to solve these questions, where an ‘incoming vergence’ ray strikes a lens surface and exits as ‘exiting vergence’:
Lv (D) + F(D) = L’v(D)
Where object incoming vergence Lv =1/-l(object distance in m) and image has exiting vergence
L’v = 1/l’(image distance in m). Since incoming ray is diverging we assign it a minus vergence power.
Here we are dealing with object distance in cm rather than meters.
Incoming Vergence L = n/-l (m)= 1.0/-l(m) is the same as = 100/-l(object distance in cm) = 100/-10cm
So L = -10.00D strikes lens surface...
Lv (D) + F(D) = L’v(D)
-10.00D + (+4.00) = -6.00 L’ (exiting image vergence in dioptres) and ray still divergent
And Image distance is the reciprocal of image vergence
L’v(D) = n/l’v(m) , so rearranging our formula for image distance l’v(m),
l’v=1/L’(D) = 1/-6.00 = -0.166m or 16.6cm
Think of this +4.00 lens surface as having plano back surface and 0 thickness. So it looks like a +4.00 lens with focal planes D =1/f, or f = 1/D = 1/+4.00 = 0.25m or 25 cm. Secondary focal place is 25 cm to right of lens and its primary focal plane is 25cm to left of lens. The object sits 10cm in front of this lens, between its primary focal plane and lens surface. Ray tracing through this (thin) +4.00D lens we find the image distance falls to left of object, is virtual and is larger. Using similar triangles to find height of image x ...
Height of Object/ Object distance = Height of image/Image distance
2cm/10cm = x/16.6cm
33.2/10 = x = 3.32cm
Magnification = height of image/height of object = 3.32/2 = 1.66X
im a bit confused what did you do with 10cm in front of a convex spherical surface how did you calculate that?
You're probably right, tmorse.
This vergence question was on a ‘thin’ plus lens. A normal ‘thick’ plus lens with index of refraction ‘n’ and centre thickness ‘t’ involves a working knowledge of principal planes and nodal points for proper ray tracing from an object to determine its image position and image size.
Then you can visualize the effect of changing the front curve (and thickness). The image size can be made larger when you increase the front (base) curve), so you can balance image size and order iseikonic lenses in situations of aniseikonia, yet you leave the power of the lens the same. Every competent optician should have an understanding of ‘geometric optics’ and the application of optical formulas to everyday dispensing.
So I suggest you find a formal opticianry school in UK, (or come to our BC College of Optics here in Vancouver, BC) and learn this and other optical formula applications. See 'BC College of Optics Sponsored Forum' below.
Doh. Good catch. I'm glad that image size ratio is proportional to before and after vergence ratio.
I'm not glad I flunk first grade arithmetic.
To be a competent optician it also helps to have a degree in psychology as you will need that as much or more than the math when dealing with many patients.
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