# Spherical mirror, shortest longest focal length

1. Mar 7, 2009

### Yroyathon

hi again. i totally rocked my physics hw this week, except for this last one.

1. The problem statement, all variables and given/known data
Consider a spherical mirror without making the paraxial approximation (Fig. 36-47, attached as a file). When a ray parallel to the axis makes an angle θ with the radius R at the point of contact, then f, the distance at which the ray crosses the axis, is given by the following.

f = R* (1 - (1 / (2*cosθ)))

Calculate the spread in values of f for a spherical mirror of radius 0.15 m and arc length 45 cm.
Shortest focal length ? cm
Longest focal length ? cm

2. Relevant equations
maybe s=r*θ

3. The attempt at a solution

i'm not totally clear on what's going on in this problem. i thought it's just a bunch of parallel rays going right and bouncing off the mirror and creating lots of different focal points. and then we just want the biggest and smallest.

so for θ, i figured that the angle between the radius (which I converted to cm) and the base line is also equal to θ. is this right?

then i used s=r * θ , and got θ = 3 radians for this particular radius and arclen. so I figured i should consider values of θ between -3/2 and 3/2, that is, so that the range is 3 radians long and split evenly between positive and negative.

plugging these values into the equation for f, i thought I was just supposed to look for the biggest and smallest values of f. but this didn't get me right answers, so... can't be that simple.

any tips for what to try next? or just conceptual help if my interpretation isn't right?

Thanks a lot.
,Yroyathon

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2. Mar 7, 2009

### Redbelly98

Staff Emeritus
Your approach seems correct. So what values did you get for f (max & min)?

Another possibility: perhaps "mirror of radius 0.15 m" refers to the actual size of the mirror, rather than the radius of curvature.

3. Mar 7, 2009

### wywong

Your argument seems ok so far. What is the maximum and minimum f you got? I think they should be 7.5cm and 0cm respectively. For $$\theta>\pi/3\approx 1.05$$, the formula yields a -ve f. What that means is that the reflected beam will meet the axis behind the mirror. It the beam is unobstructed, it will hit the mirror once more before meeting the axis somewhere in front of the mirror, resulting in a +ve f.

4. Mar 7, 2009

### Yroyathon

the first time I tried 7.5 cm min, and 22.5 cm max. both were wrong.

second time i tried 0 cm min and 91.0262 cm max. both were wrong.

i don't have my notes here, but i'll comment again tomorrow after i've tried to incorporate what you've commented.

but given my values of what's wrong, feel free to further comment if these bad figures are helpful.

thanks!

5. Mar 7, 2009

### Redbelly98

Staff Emeritus
7.5 cm works for θ=0, R=15 cm

However, something is wrong. When you plug in 1.5 radians, you should get something other than what you are saying. Can you show your calculation for the 1.5 radians case?

6. Mar 8, 2009

### Yroyathon

still haven't gotten this one. but here are the two graphs of that function that I'm considering. from -1.5 to 1.5 for theta values, and from 0 to 3 for theta values.

thoughts?

well, i'm pretty much out of time anyway, but i'd be interested in what you folks thinks.

thanks much.

,Yroyathon

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7. Mar 8, 2009

### Yroyathon

nevermind, i just got it.

apparently f_max was 7.5 (as the one graph shows), and the min is literally the -91.0262. it was my impression that this negative focal length should just be turned into a pos length, but i guess the problem literally wanted the max and min vals of this function over that specific domain? huh.

thanks for the help everybody, this place is awesome.

8. Mar 9, 2009

### Redbelly98

Staff Emeritus
You're welcome, glad you got it to work out.

But I take issue with whoever came up with this problem. As wywong pointed out, the rays would simply reflect from the mirror a 2nd time, and intersect at some point of positive focal length.