Assignment 18

course Phy 232

018. `Query 16*********************************************

Question: univ phy problem 34.86 (35.52 10th edition) f when s'=infinity, f' when s = infinity; spherical surface.

How did you prove that the ratio of indices of refraction na / nb is equal to the ratio f / f' of focal lengths?

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Your solution:

na/s+nb/s’=(nb-na)/R

s = na/((nb-na)/R - nb/s’)

and

s’ = nb/((nb-na)/R - na/s)

if you take the limit as s and s’ go to infinity, then you get

f=-na*R/(na-nb)

and

f’=-nb*R/(na-nb)

then if you divide f by f’, then you get

f/f’ = na/nb

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Given Solution:

`a** The symbols s and s' are used in the diagrams in the chapter, including the one to which problem 62 refers. s is the object distance (I used o in my notes) and s' the image distance (i in my notes). My notation is more common that used in the text, but both are common notations.

Using i and o instead of s' and s we translate the problem to say that f is the object distance that makes i infinite and f ' is the image distance that makes o infinite.

For a spherical reflector we know that na / s + nb / s' = (nb - na ) / R (eqn 35-11 in your text, obtained by geometrical methods similar to those used for the cylindrical lens in Class Notes).

If s is infinite then na / s is zero and image distance is s ' = f ' so nb / i = nb / f ' = (nb - na) / R.

Similarly if s' is infinite then the object distance is s = f so na / s = na / f = (nb - na) / R.

It follows that nb / f ' = na / f, which is easily rearranged to get na / nb = f / f'.

THIS STUDENT SOLUTION WORKS TOO:

All I did was solve the formula:

na/s+nb/sprime=(nb-na)/R

once for s and another time for sprime

I took the limits of these two expressions as s and s' approached infinity.

I ended up with

f=-na*r/(na-nb)

and

fprime=-nb*r/(na-nb)

when you take the ratio f/fprime and do a little algebra, you end up with

f/fprime=na/nb **

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Self-critique (if necessary):

I got the question correct but had to look at the answer for help on the problem

Self-critique Rating:

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Question: `q **** univ phy How did you prove that f / s + f' / s' = 1?

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Your solution:

From previous problem,

f/f’ = na/nb

f=na*R/(na-nb)

f’=nb*R/(na-nb)

na/s+nb/s'=(nb-na)/R

R * na /[s (na - nb)] + R * nb / [s ' ( na - nb)]= 1.

Subsitutig in f and f’, we get f/s + f’/s’ = 1

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Given Solution:

`a** We can do an algebraic solution:

From nb / f ' = (nb - na) / R, obtained in a previous note, we get f ' = nb * R / (nb - na).

From na / f = (nb - na) / R we get f = na * R / (nb - na).

Rearranging na/s+nb/s'=(nb-na)/R we can get R * na / ( s ( na - nb) ) + R * nb / (s ' ( na - nb) ) = 1.

Combining this with the other two relationships we get f / s + f ' / s / = 1.

An algebraic solution is nice but a geometric solution is more informative:

To get the relationship between object distance s and image distance s' you construct a ray diagram. Place an object of height h at to the left of the spherical surface at distance s > f from the surface and sketch two principal rays. The first comes in parallel to the axis, strikes the surface at a point we'll call A and refracts through f ' on the right side of the surface. The other passes through position f on the object side of the surface, encounters the surface at a point we'll call B and is then refracted to form a ray parallel to the axis. The two rays meet at a point we'll call C, forming the tip of an image of height h'.

From the tip of the object to point A to point B we construct a right triangle with legs s and h + h'. This triangle is similar to the triangle from the f point to point B and back to the central axis, having legs f and h'. Thus (h + h') / s = h / f. This can be rearranged to the form f / s = h / (h + h').

From point A to C we have the hypotenuse of a right triangle with legs s' and h + h'. This triangle is similar to the one from B down to the axis then to the f' position on the axis, with legs h and f'. Thus (h + h') / s' = h / f'. This can be rearranged to the form f' / s' = h' / (h + h').

If we now add our expressions for f/s and f'/s' we get

f / s + f ' / s ' = h / (h + h') + h' / (h + h') = (h + h') / (h + h') = 1.

This is the result we were looking for. **

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Self-critique (if necessary):

I did not do as an in depth answer as the solution, but I got the overall answer correct.

Self-critique Rating:

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&#This looks good. Let me know if you have any questions. &#