Question

http://assets.openstudy.com/updates/attachments/56ee157be4b0e02a5e0c09b7-ganeshie8-1458443662346-dd.png

Answer 1

Are you aiming to become an optometrist lol

Answer 2

lol just randomly picked this problem as it looks interesting

Answer 3

I think part a of the question is asking this :

Give that for seeing objects at infinity f = 2.5cm, find out the f for seeing objects at distance p = 40cm

Answer 4

brb

Answer 5

I'd like to get more into this since I see you talking about it a lot, I am looking online to find some derivations/experiments/explanations of snell's law, indices of refraction, and the lensmaker equations.

Answer 6

okay sure

\(\dfrac{1}{\infty} + \dfrac{1}{i} = \dfrac{1}{f}\)

if im interpreting this correctly, the distance between cornea and retina must be \(i=f=2.5cm\)

Answer 7

@Kainui yeah this chapter is based on those two laws :

1) reflection law : angle of incidence = angle of reflection
2) snells law : \(n_1\sin\theta_1 = n_2\sin\theta_2\)

Answer 8

The reflection law seems reasonable but snell's law I'm not sure how I could see that. Is there some intuition or cute example or derivation so I don't mistakenly think it's, say, \(n_1 \cos \theta_1 = n_2 \cos \theta_2\)?

Answer 9

you have to find an f'

Answer 10

i think snell's law is telling us how the speed of light changes in different media

\(n_1sin\theta_1\)
here \(n_1 = \text{index of refraction} = \dfrac{c}{v}\)
\(c\) = speed of light in vacuum
\(v\) = speed of light in the medium

Answer 11

Yes, but you don't have to use this, you have to find a relation for the images formation

Answer 12

Just for the sake of simple analysis, replace \(\sin\theta\) by \(\theta\) and the snells law becomes :
\[n_1\theta_1 = n_2\theta_2\]

If \(n_2\gt n_1\), then \(\theta_2\lt \theta_1\)
In otherwords, light bends more as at the interface as it passes to a medium with greater index of refraction

Answer 13

This is intuitively pleasing if with the four wheeler analogy.


http://www.2020mag.com/nysso/NYSSOExams/107207/Nature%20of%20Light11-00.jpg

Notice how the car bends as it changes medium because of the difference in distance between left and right wheels

Answer 14

I'm trying to think about some concrete examples of this, like I know when you look into water stuff looks like it bends. I vaguely recall \(n_{air} \approx 1\) but what about water or other materials I guess I should look that up.

ok I like this picture

Answer 15

\[(1/s')-(1/s)=1/f' \]

Answer 16

yeah mira back to the original problem

so 2.5cm is the image distance

Answer 17

yep

Answer 18

Oh then its just algebra is it

Answer 19

yes

Answer 20

\[\dfrac{1}{40} + \dfrac{1}{2.5} = \dfrac{1}{f'}\]

Answer 21

Correct

Answer 22

f' = 2.35
http://www.wolframalpha.com/input/?i=1%2F%281%2F40+%2B+1%2F2.5%29

Answer 23

eheheehehee

Answer 24

it seems the focal length of the eye decreases as the object moves from infinity towards the eye

Answer 25

Yep. so this is why we don't use the senslls law because we don't think in the moment that light is in the lens, only in the case when the light passes across the lens, right?

Answer 26

gotcha! 1/p+1/i = 1/f is derived from the two basic laws though
part b seems to be a qualitative type..

Answer 27

You use the senlls law when you have a problem in which the light gets into another medium, like water of if you want to study the behaviour of the light into the lens or a glass, but for our problem we don't mind what happens with the light when it goes into the lens

Answer 28

Yep :)

Answer 29

Just work! For no reason at all!
Well except for the reason to expand your knowledge!
My type of reasons!

Answer 30

makes sense...

Answer 31

Love to help but I have my own problems right now! Calculus and Physics got me tied up. lol

Answer 32

We're good comrad :^)

Answer 33

so the muscles attached to the lens of eye have this task of reducing the focal length of lens from 2.5 to 2.35

Answer 34

need to relate focal length to the curvature of the lens hmm

Answer 35

\[1/f'= (n-1)(1/R1-1/R2)\]
In this case n is the refraction index of the lens
R1 and R2 are the radius curvature

Answer 36

\(f\propto r\)

?

Answer 37

then the muscles need to contract the lens ?

Answer 38

So that equation makes the relation between the radius curvature and the f'

Answer 39

think about the f' that you got and how the R1 and R2 will change... if the f'get bigger or smaller, what could happen?

Answer 40

assuming the same radius of curvature for both sised we have :
r1=r, r2=-r, the equation becomes :

1/f'= (n-1)(2/r)

Answer 41

if f decreases, r has to decrease i think

Answer 42

^smart people

Answer 43

but you have to substact, so, you got a cero in hat case. 1/f'=(n-1)(1/E1-1/R2)
if R1 = R2 we have a cero

Answer 44

r1 = -r2

Answer 45

but you got the answer, if f' decreases R will decrease.