optics conceptual question

[tank you]

lets say were talking about a telescope made of glass converging lenses.

why would the focal length decrease if the glass was replaced with a material having a greater index of refraction?

similarly, what would happen to the magnification if the air within the telescope was replaced with water?

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[cfdavid]

The higher the index of refraction, the more the light will bend. You can relate this to farsightedness in that the goal is to shorten the focal length of the object that you are looking at. Many farsighted people have shorter length eyeballs. So, the retina is closer to the lens/cornea. Therefore, in order to get the focal point to 'hit' the retina, they will need convex/converging lenses in their glasses. This helps bend the incoming light more, so that the light rays converge on the retina and not behind it.

Regarding the microscopes, relating Power to focal length P=1/f Therefore, the shorter the focal length, the more 'powerful' the lens.

 

[BaylorGuy]

As stated above...the greater the index of refraction of the lens, the greater the bend of the incoming light. Assume a parallel light ray hitting near the top of a glass lens, the light will bend and hit the focal point. now assume the same light ray hitting near the top of a diamond lens (or a lens with a higher index of refraction), the light will bend moreso than it did with the glass, thus the focal length will be shorter than the previous focal length. Since this is the case, due to P=1/f the power of the lens increases...(it will generally increase with increasing index of refractions).

As for your second question, i'm not totally positive, although i will try to reason it. the incident light will bend more if it goes from glass to air (lens to focal), than if it does from glass to water (lens to focal). Since the light through water will not bend as strong as it does through air, the focal length will increase. If the focal length increases, the image distance will increase. Since magnification=-(image distance/object distance), and increase in image distance (due to increasing focal length), will lead to an increase in magnification.

From i can tell, that should be right, but if anyone thinks its wrong, call me out.

 

[Mediculous]

This obviously is more lengthy than just knowing what happens, but here's how you mathematically prove the answers:

di= image distance do= object distance xi= incident angle xr=refracted angle
f= focal length
ng, nu, na, and nw are the indices of refraction for glass, unknown, air, and water, respectively.

First Question- Snell's Law: ng x sin(xi)= na x sin(xr)
nu>ng and na is constant, so the right side of
the equation can only increase if you increase sin(xr) which means xr increases.

Second Question (1/di) + (1/do)= (1/f)
di= (do x f)/(do-f)
do is a constant in this instance.
Using Snell's Law, you get an increase in f in this instance.(see above)
y is the factor of increase of the original focal length. and fy>f1 so (do x yf)/(do-yf) > (do x f1)/(do-f1) because
y(do x f)/(do-yf) = (y(do-f))/(y((do/y)-f))
cancel out y and the (do/y)-f is always smaller than do-f, and since your dividing by a smaller number, you get a larger number. And then this larger di gives a larger magnification.

Where's Shrike to explain this more eloquently?

 

[ravin]

I haven't had all of optics, but:
if light passes through air to water will the light be refracted Away from the normal?

 

[Mediculous]

I haven't had all of optics, but:
if light passes through air to water will the light be refracted Away from the normal?

Yes, unless the angle of incidence is normal to the water.
n1 x sin(angle1) = n2 x sin(angle2) Since the indices of refraction can not equal zero, if the one of the angles is zero, so is the other.(Technically, one of the angles could be 180 degrees, but that means there is reflection instead of refraction)

 

[BaylorGuy]

general rule....air has index of refraction almost equal to 1 (only vaccuum is less). any incident ray of light going from LOWER index to HIGHER index will bend TOWARD the normal.

HIGHER index to LOWER index, will bend AWAY from normal.

light will bend toward normal if going from air to water.

 

[Mediculous]

Sorry, when I read your question I was thinking that by "Away from the normal" you meant any deviation from the normal or if there might be total internal reflection even. The capital "A" should have cued me in more. BaylorGuy is right.

 

[tank you]

This obviously is more lengthy than just knowing what happens, but here's how you mathematically prove the answers:

di= image distance do= object distance xi= incident angle xr=refracted angle
f= focal length
ng, nu, na, and nw are the indices of refraction for glass, unknown, air, and water, respectively.

First Question- Snell's Law: ng x sin(xi)= na x sin(xr)
nu>ng and na is constant, so the right side of
the equation can only increase if you increase sin(xr) which means xr increases.

Second Question (1/di) + (1/do)= (1/f)
di= (do x f)/(do-f)
do is a constant in this instance.
Using Snell's Law, you get an increase in f in this instance.(see above)
y is the factor of increase of the original focal length. and fy>f1 so (do x yf)/(do-yf) > (do x f1)/(do-f1) because
y(do x f)/(do-yf) = (y(do-f))/(y((do/y)-f))
cancel out y and the (do/y)-f is always smaller than do-f, and since your dividing by a smaller number, you get a larger number. And then this larger di gives a larger magnification.

Where's Shrike to explain this more eloquently?

im sure that explains it, now if only i knew what u were talking about....

 

[Shrike]

Where's Shrike to explain this more eloquently?

Again, the pressure. But I am going with BaylorGuy's explanation of this problem. Much better to think conceptually than to dive into equations. (Your equations look right, by the way, but at this point people shouldn't be trying to hit that sort of thing too hard, in my opinion.)