Physics lenses and mirrors question- PLEASE CLARIFY

[ptavasso]

Hi everyone,

This is my understanding:

CONCAVE lens- diverges; smaller, virtual image (upright or inverted??)

CONVEX lens- converges; larger, real image (upright or inverted??)

CONCAVE mirror- converges; smaller, real image (upright or inverted??)

CONVEX mirror- diverges; larger, virtual image (upright or inverted??)

 

1. A real image is always inverted and virtual image is always upright, correct?

2. Why don't the reflected rays from a concave lens form the same image as a concave mirror? Or do they and it's just not as prominent because very little rays are reflected?

3. A concave lens forms a virtual image because it's rays never meet (diverge from each other), correct?

4. An object is always on the same side as the eye for a mirror (because you stand in front of a mirror). An object is on the opposite side as the eye for a lens (you look at an object through a lens). Is this correct? (I'm not talking about which side the image is on)

5. An object is + for both lens and mirror; the eye is + for mirror and – for lens. Is this correct?

 

This isn't really related to the above, but can you check if this is correct:

Shorter wavelength = higher frequency = more energy = bend more (I edited this after someone pointed out my typo!)

If so, why does more energy lead to more bending? I know that more energy means a higher refractive index, but I don't understand this- please explain why in simple, practical terms.

 

 

Please don't explain laws or equations. I've searched the internet and textbooks to clarify and check my understand and all that provided was equations and laws so please simply read my description and comment on it.

Thank you!

[Futbol99]

Hi everyone,

This is my understanding:

a. CONCAVE lens- diverges; smaller, virtual image (upright or inverted??)

b. CONVEX lens- converges; larger, real image (upright or inverted??)

c. CONCAVE mirror- converges; smaller, real image (upright or inverted??)

d. CONVEX mirror- diverges; larger, virtual image (upright or inverted??)

 

Im going to try to answer your questions to the best I can as Im also studying this rite now for the mcat, but other help/discussion would be well, just incase Im wrong as Im not 100%

 

 

a. Concave Lens = diverges, virtual, UPRIGHT

b. Convex lenx = Converges, real image, INVERTED

c. Concave mirror = converges, real image, INVERTED

d. Convex mirror = diverges, virtual image, UPRIGHT

 

The sizes (magnification) that you mention seem right. The way it looks right to me is using the magnification equation (m = -i/o); where i = image and o= object distance

 

a real image means i > 0 AND virtual image means i < 0. If you look at the magnification equation; m= - i /o (image over object distance). So for a real image; (i > 0) makes m<0 (thus maginifaction would be smaller than original) and for a virtual; i < 0, m>0 and so magnification would be larger than original. (that tells us about our upright/inverted position which answers #1 as well: (real image is i>0, means m<0, which means real image is ALWAYS inverted, and virtual is upright)

 

Sizes are such as Im assuming object distance doesn't change (which is the o in the magnification equation I mentioned...that could cause difference in the magnification, but note that o is ALWAYS positive, so it all depends on your image whether its real or virtual).

 

2. A mirror and a lens are completely different things. A mirror REFLECTS and a lens REFRACTS. So this is why a concave mirror will not do the same thing as a concave lens, they're different structures. A concave mirror will reflect an inverted, real image, while a concave lens diverges, and produces a virtual, upright image.

 

3. Correct. I think its just more proper to say that the rays intersect towards a (real) focal point. (whereas for convex mirrors...rays intersect at the (imaginary) focal point so yeah i.e, they don't intersect)

 

4/5- sorry not really sure about this... all I know is that the object is always a positive distance away from the mirror/lens. So in front of a mirror, it would be front.

 

As for the energy bending stuff...I'm not really positive but I will see if I can apply what I know. (Im also studying this, and maybe someone else can put in their opinion).

For one, Im not sure where you got Less energy = more bending? Im really interested so I know about this concept properly...

Im going to try explaining using the concept of dispersion and Power of lens.

- In the Dispersion concept, it basically shows us about the bending of light (refraction) of 2 different materials that vary with wavelength. It states that a material with shorter Wavelength, higher energy, smaller Vmedium, and thus Higher refractive index will bend MORE (refract more).

 

So for example looking at violet light vs. red light, Violet light is shown to refract MORE than Red. Violet has a SHORTER wave and thus MORE energy (than red). Because violet has shorter Wavelength, its proportional to V (through the v =(wavelength)(frequency)) thus V is smaller as well. Small V means HIGH (n) - refractive index. So a higher energy molecule, smaller wavelength, smaller V, higer n should have a greater refraction. (so this contradicts what you were saying above :S)

 

- This dispersion concept is consistent with something more proper to explain lens, which talks about the property of the lens. Power of lens equates to (P = 1/f) f is FOCAL POINT.

So basically a lens with HIGH power of refraction lens (a THICKER lens), can refract light to a smaller focal point (which is the point at which light rays refract right after passing through convex lens) so the smaller the focal point that the light rays refract at, and more bending.

 

(Power could be related to energy proportionally, and so a lens that has more energy can reflect more light closer (but I dont think thats really important relationship discussing lens)

 

So in short from what I was meaning to say:

- I think talking about LIGHT properties, higher energy, and higher refractive index would cause more bending (closer to the lens) vs. something of lower energy.

 

- But more properly, I think that light bending through a lens is better explained by the power of the lens it self. A thicker lens would bend any light with the same properties more than a thin lens and thus shows the difference in power of lens (and their size - thick vs thin.)

 

NOT entirely sure... lol. Hope this helps some way, and please please correct me if I went anywhere wrong, I have my mcat coming up as well, and optics is probably my least comfortable section lol.

[ptavasso]

Thanks a lot!

I'm glad you confirmed most of my points

 

I completely messed up in my question, you are correct that shorter wavelength = more bending/ more diffraction. Sorry I mistyped and didn't proofread (oops)!

So...

shorter wavelength = higher frequency = more energy = bend more

 

As well, I still need clarification on:

4. An object is always on the same side as the eye for a mirror (because you stand in front of a mirror). An object is on the opposite side as the eye for a lens (you look at an object through a lens). Is this correct? (I'm not talking about which side the image is on)

5. An object is + for both lens and mirror; the eye is + for mirror and – for lens. Is this correct?

[Futbol99]

Thanks a lot!

I'm glad you confirmed most of my points

 

I completely messed up in my question, you are correct that shorter wavelength = more bending/ more diffraction. Sorry I mistyped and didn't proofread (oops)!

 

 

 

Haha, no problem, sounds good to me! Glad I could help. Sorry I couldnt help with 4/5, hoping someone can contribute an answer for the both of us!