Optics question - images and a glass rod with curved ends Pearson 34_76

[rogue_system]

Homework Statement

Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is 6.00 cm, and the radius of curvature at the right end is 12.0 cm. The length of the rod between vertices is 39.0 cm . The object for the surface at the left end is an arrow that lies 24.0 cm to the left of the vertex of this surface. The arrow is 1.80 mm tall and at right angles to the axis.

What is the object distance for the right surface?

Relevant Equations

n1/s + n2/s' = (n2-n1)/R

n1=1.00 (air), n2=1.60 (glass), R=6.00cm, s=24cm

(1/24) + (1.60/s') = (1.60-1.00)/6

Simplifying I get

[(0.60/6)-(1/24)]*1.60=1/s'=7/75

s'=75/7

This puts s' to the left hand (same side as incoming light) of the lens on the left hand side, so the distance to the lens on the right hand side would just be L+s'? For that I get 39+(75/7) which gives us about 43.412cm, but that is apparently wrong.

I've written this out with my tablet to make the formatting easier to read, and also a drawing in case I am visualizing this wrong.

There's more parts later on, but I am already stuck.

 

[rogue_system]

Just saw that I multiplied when I should've divided, so I might just need a nap.

Simplifying should be:

[(0.60/6)-(1/24)]/1.60=1/s'=17/192

Which gives:

s' = 192/17 or about 11.29cm

Adding to L to find S2 would give about 50.294cm, but I'm still not sure I've done that right? Corrected image attached.

 

[kuruman]

Your drawing is incomplete. What happens to a ray that leaves the tip of the arrow and moves to the right parallel to the optical axis? I suggest that you make a drawing to scale and consider whether the ray from the tip of the arrow exits before or after it reaches the side of the rod.

 

[rogue_system]

Your drawing is incomplete. What happens to a ray that leaves the tip of the arrow and moves to the right parallel to the optical axis? I suggest that you make a drawing to scale and consider whether the ray from the tip of the arrow exits before or after it reaches the side of the rod.

Not sure how accurate my scale is, but I drew it larger to better visualize. Since the lens is convex I would assume that the image would be inside, but I got a positive value for S' so I calculated it as though it was on the same side as S. It's entirely probable that I have misunderstood the sign convention for S' though.

If S' being positive puts it to the right of the optical surface, then the diagram should look like the below and then S2 would be L-S'?

 

[kuruman]

Is there a figure that comes with this problem? If so, please post a screenshot. Specifically, what is the diameter of the rod?

 

[rogue_system]

Is there a figure that comes with this problem? If so, please post a screenshot. Specifically, what is the diameter of the rod?

Thre is not, and that information is not given. What I posted in the original problem statement is all it gives.

 

[kuruman]

Thre is not, and that information is not given. What I posted in the original problem statement is all it gives.

In that case this is the best that can be done on the basis of the information provided in the statment of the problem. Note that "hemispherical" surfaces at the two ends are literally half-spheres and that the separation between ends (assumed to be the "vertices") are 39 cm apart. Draw a ray diagram of the ray from the tip of the arrow which is at left end of the axis but too small to be drawn to scale.

Personally, I don't think that this problem is instructive. Where did you find it?

On edit
This figure is probably not what the author meant. See post #12.

 

[TSny]

Not sure how accurate my scale is, but I drew it larger to better visualize. Since the lens is convex I would assume that the image would be inside, but I got a positive value for S' so I calculated it as though it was on the same side as S. It's entirely probable that I have misunderstood the sign convention for S' though.

Review the sign conventions associated with the equation $\frac{n_1}{s} + \frac {n_2} {s'} = \frac{n_2 - n_1}{R}$.

If S' being positive puts it to the right of the optical surface, ... then S2 would be L-S'?

Yes, that's right. Check that this is correct for both $S' < L$ and $S' > L$. You'll need to know the sign conventions.

 

[TSny]

In that case this is the best that can be done on the basis of the information provided in the statment of the problem. Note that "hemispherical" surfaces at the two ends are literally half-spheres and that the separation between ends (assumed to be the "vertices") are 39 cm apart. Draw a ray diagram of the ray from the tip of the arrow which is at left end of the axis but too small to be drawn to scale.

View attachment 374749
Personally, I don't think that this problem is instructive. Where did you find it?

Interesting. I did not interpret the problem this way. But I see now that the problem does state "hemispherical."

@rogue_system : Make the sign correction mentioned above in post #8 and see if you get the expected answer.

 

[kuruman]

Interesting. I did not interpret the problem this way. But I see now that the problem does state "hemispherical."

That is why I asked OP to provide a figure.

 

[renormalize]

Personally, I don't think that this problem is instructive. Where did you find it?

Variations of this problem appear in lots of the online paid problem-solution sites like Chegg, Filo, etc. Here's one example from Vaia:

 

[kuruman]

I suppose then that "hemispherical surface" is a misnomer. We have a cylindrical rod of unspecified diameter capped by spherical segments of different radii that are coaxial with the cylinder. This makes more sense.

 

[TSny]

I suppose then that "hemispherical surface" is a misnomer. We have a cylindrical rod of unspecified diameter capped by spherical segments of different radii that are coaxial with the cylinder. This makes more sense.

Yes. I think either interpretation would yield the same answer.

The top figure has the full hemispheres. The bottom figure is a cylinder cut out of the top figure.

The height of the object is small compared to the radii of the hemispheres, and we assume paraxial rays. So, the rays traveling in the top figure will remain inside the cylindrical portion.

 

[rogue_system]

Yes. I think either interpretation would yield the same answer.

View attachment 374753

The top figure has the full hemispheres. The bottom figure is a cylinder cut out of the top figure.

The height of the object is small compared to the radii of the hemispheres, and we assume paraxial rays. So, the rays traveling in the top figure will remain inside the cylindrical portion.

The smaller cylinder with curved ends matching the radius of the curvature is more or less how I pictured it (basically a camera lens without the metal barrel around it). Switching the sign got me to the correct distance for S' and thus $S_{2}$.

I had switched S to be negative on the left hand side because of the 'similar' explanation given by Pearson here. (The equation they use isn't rendering properly on my monitor, I am more and more understanding the importance of LaTeX, and will try use it more often.)

Since the image is still inside the optical system, even though the lens on the right hand side hasn't been accounted for, I guess it's considered a "real" instead of "virtual" image? If the optical system were shorter than S', and the image created by the left hand side fell outside of the optical system would it be considered virtual?

Edit: There is apparently something I am still not understanding, trying to figure out what S' is for the right hand side of the optical system.

$$\frac {1.60} {S_{2}} + \frac 1 {S'_{2}} = \frac {1-1.6} {-12}$$

Which gives me about 9.23cm for ${S'_{2}}$ which is apparently incorrect?

Disregard, I've gone around in circles enough that I mixed up S' and ${S_{2}}$ in my calculations.

 

[kuruman]

The height of the object is small compared to the radii of the hemispheres, and we assume paraxial rays. So, the rays traveling in the top figure will remain inside the cylindrical portion.

Still, doesn't one have to trace the paraxial ray and find the point of incidence on the right spherical surface to determine the size of the cylinder?

 

[TSny]

Still, doesn't one have to trace the paraxial ray and find the point of incidence on the right spherical surface to determine the size of the cylinder?

Yes. I see your point.

For this problem, a sketch shows that the rod's diameter would have to be very small to cause a problem.

Here is a diagram that I tried to draw to scale.

Each small square has a side of 1 cm. The object is given to be only 0.18 cm tall. But I drew it 1 cm tall to make it easier to see. So, the object is actually only about 1/5 the size shown.

I drew only one ray from the tip of the object, parallel to the optical axis. It passes through the focal point, F, associated with the left end of the rod. Calculation shows that F is 16 cm to the right of the left end and the image of the left end is 27.4 cm to the right of the left end. You can see that it strikes the right end of the rod at about 1.4 cm below the axis. If I had drawn the object 0.18 cm tall, then the ray would hit the right end at approximately $\frac{1}{5} \cdot 1.4$ cm $\approx 3$ mm below the axis. So, the rod only needs to have a diameter greater than about 6 mm. A rod 6 mm in diameter and 39 cm long would be a very thin rod.

 

[kuruman]

If I had drawn the object 0.18 cm tall, then the ray would hit the right end at approximately 15⋅1.4 cm ≈3 mm below the axis

I agree, I got 2.6 mm below the axis.