How does the index of a medium affect the focal length of a lens?

[physicsphreak2]

Homework Statement

Suppose we have a lens of known focal length, f=100 cm.
(a) If light enters the lens from air, but after the lens enters a medium with index 1.33, how far from the lens will focus? (Assume the incident light is collimated.)
(b) What if there is instead air after the lens, except for a 5 cm patch of medium with index 1.5?

Homework Equations

1/f=(n-1)(1/R1-1/R2)

The Attempt at a Solution

I don't have the index of the lens, so maybe I don't need to use the lens-maker's equation?

So instead, I proceed like this: Let's say the light travels distance L in air, and x in the medium. Clearly L+x=f because you travel through the two mediums until you got to the focus. If L+x=f, but some of x is replaced by something of effective length x/n, then you have to travel L=f-(x/n) in the air still. So the total travel distance is L+x=(f-(x/n))+x= f+x(1-1/n).

For (a), then, we get that the light focuses to a point at 100+100(1-1/1.33)=124.8 cm

For (b), then, we get that the light focuses to a point at 100+5-(5/1.5)=101.67 cm.

This makes sense, because I would expect a lens to be less powerful in a higher index medium... but is this right?Also, I'm confused about whether I would DIVIDE by the index (since really we care about diffraction effects, and those are proportional to wavelength, which gets smaller in a medium) or MULTIPLY by the index (because that's what happens to the optical path length, although I don't see why we care about the optical path length for focusing, isn't that more for interference effects?)

 

[Redbelly98]

Homework Statement

Suppose we have a lens of known focal length, f=100 cm.
(a) If light enters the lens from air, but after the lens enters a medium with index 1.33, how far from the lens will focus? (Assume the incident light is collimated.)
(b) What if there is instead air after the lens, except for a 5 cm patch of medium with index 1.5?

Homework Equations

1/f=(n-1)(1/R1-1/R2)

The Attempt at a Solution

I don't have the index of the lens, so maybe I don't need to use the lens-maker's equation?

So instead, I proceed like this: Let's say the light travels distance L in air, and x in the medium. Clearly L+x=f because you travel through the two mediums until you got to the focus. If L+x=f, but some of x is replaced by something of effective length x/n, then you have to travel L=f-(x/n) in the air still. So the total travel distance is L+x=(f-(x/n))+x= f+x(1-1/n).

For (a), then, we get that the light focuses to a point at 100+100(1-1/1.33)=124.8 cm

For (b), then, we get that the light focuses to a point at 100+5-(5/1.5)=101.67 cm.

This makes sense, because I would expect a lens to be less powerful in a higher index medium... but is this right?Also, I'm confused about whether I would DIVIDE by the index (since really we care about diffraction effects, and those are proportional to wavelength, which gets smaller in a medium) or MULTIPLY by the index (because that's what happens to the optical path length, although I don't see why we care about the optical path length for focusing, isn't that more for interference effects?)

Welcome to Physics Forums. Diffraction and interference are irrelevant here.

For (a), I get something slightly different -- and simpler. I sort of follow your reasoning, but am not convinced it is valid.

Try a simple geometric approach. Light leaves the lens and, if there is just air, focuses to a point a distance f away. So consider the right triangle that is formed by:

• the central axis, from the lens center to the focal point
• a ray drawn from the top of the lens to the focal point
• a line drawn from the lens center to the top of the lens

Now replace the air with the medium of index 1.33. If you like, imagine a very thin layer of air between the lens and the medium. So the ray emerging from the top of the lens, instead of making an angle θ with the horizontal, makes an angle ___(?) with the horizontal. Figure out that angle, then use geometry to figure out where it reaches the central axis.

And use the usual small-angle approximations.

Hope that helps.

 

[Basic_Physics]

Using Redbelly's suggestion you can choose your own refraction angle of the beam coming off at the top of the lens towards the focal point in air. Then using Snell's law figure out the perpendicular using f = 100 cm from this angle. Then again apply Snell's law with the beam now refracting out into the new medium.