- #1
w2dot
- 1
- 0
Problem 1
A goldfish's eye is 4cm from the surface of a spherical goldfish bowl of radius 10cm. Neglecting the thickness of the glass, find the apparent position and linear magnification of the eye to an observer if the refractive index of the water is 1.333. (Hint: have the light traveling from left to right to meet the surface of the bowl)
From "Optics" 2nd edition 1996 by AH Tunnacliffe and JG Hirst. Pg 62, Question 20
(1) Lreduced' - Lreduced = F
(2) Lreduced = n/l L'reduced=n'/l'
(3) h'/h = Lreduced/L'reduced
(4) (n'-n)/r = F
Using (1) and (2)
n'/l' - 1/l = (1-1.333)/(-0.1)
L' - 1.333/(-0.04) = 3.33
L' = -29.95
l' = -0.03334m
Magnification
L/L' = (1.333/-0.04)/(-29.95) = 1.11
That gives me the right answers, but I'm really confused why that is.
Should I do this problem with r = +0.04 or -0.04?
Why, for L, would I treat n as 1 when it's on the side of the water?
Why doesn't n1/n2 = real depth/apparent depth work?
Problem 2
A spherical refracting surface, separating air from glass, forms a real image twice the size of the real object. If the image is 6 times as far from F' as the object is from F', find the refractive index of the glass.
From same source as above but Q24
xx' = ff'
h'/h = Lbar/Lbar'
F = -n/f
F = n'/f'
L' - L = F
h' = 2h2h/h = Lbar/Lbar'
2 = Lbar/Lbar'
Lbar=1/l (because in air)
l = x+f
Lbar' = n'/l'
l' = 6x + f' (because image's x is 6x object's x)
2 = Lbar/Lbar'
2 = (1/(x+f))/(n'/(6x+f'))
2 = (6x + f')/((x + f)n')
2n'x + 2n'f = 6x + f'
This is where I wonder whether the f's are equal and if not, what they are. I've found introducing r from F=(n'-n)/r (as f = -n/F and f' = n'/f') makes it even more muddled
THANK YOU!
Homework Statement
A goldfish's eye is 4cm from the surface of a spherical goldfish bowl of radius 10cm. Neglecting the thickness of the glass, find the apparent position and linear magnification of the eye to an observer if the refractive index of the water is 1.333. (Hint: have the light traveling from left to right to meet the surface of the bowl)
From "Optics" 2nd edition 1996 by AH Tunnacliffe and JG Hirst. Pg 62, Question 20
Homework Equations
(1) Lreduced' - Lreduced = F
(2) Lreduced = n/l L'reduced=n'/l'
(3) h'/h = Lreduced/L'reduced
(4) (n'-n)/r = F
The Attempt at a Solution
Using (1) and (2)
n'/l' - 1/l = (1-1.333)/(-0.1)
L' - 1.333/(-0.04) = 3.33
L' = -29.95
l' = -0.03334m
Magnification
L/L' = (1.333/-0.04)/(-29.95) = 1.11
That gives me the right answers, but I'm really confused why that is.
Should I do this problem with r = +0.04 or -0.04?
Why, for L, would I treat n as 1 when it's on the side of the water?
Why doesn't n1/n2 = real depth/apparent depth work?
Problem 2
Homework Statement
A spherical refracting surface, separating air from glass, forms a real image twice the size of the real object. If the image is 6 times as far from F' as the object is from F', find the refractive index of the glass.
From same source as above but Q24
Homework Equations
xx' = ff'
h'/h = Lbar/Lbar'
F = -n/f
F = n'/f'
L' - L = F
The Attempt at a Solution
h' = 2h2h/h = Lbar/Lbar'
2 = Lbar/Lbar'
Lbar=1/l (because in air)
l = x+f
Lbar' = n'/l'
l' = 6x + f' (because image's x is 6x object's x)
2 = Lbar/Lbar'
2 = (1/(x+f))/(n'/(6x+f'))
2 = (6x + f')/((x + f)n')
2n'x + 2n'f = 6x + f'
This is where I wonder whether the f's are equal and if not, what they are. I've found introducing r from F=(n'-n)/r (as f = -n/F and f' = n'/f') makes it even more muddled
THANK YOU!
Last edited: