Geometrical optics- position of image

In summary: But you told me that n1=1.33 and n2=1 (post #1) ... so where did that 0.33 come from, and why is n1 first?What you wrote there does not match your notes, and nor does it match the link I gave you.surely: n2 - n1 = 1 - 1.33?Try doing the algebra one line at a time. Be super careful: you appear to have a blind spot there.but suprisingly if I take r = +0.12 , i got the ans which is 0.072m . by taking r= -0.12
  • #1
desmond iking
284
2

Homework Statement



A sphere of radius 12.0cm has refractive index of 1.33. A speck is 4.0cm from the centre of the sphere is viewed along the diameter that passes thru the speck. Find the position of image when the speck is viewed from the nearer side?
the ans is -7.2cm

Homework Equations


The Attempt at a Solution


n1= 1.33 n2= 1 , u = +0.08m. Assuming light travel form right to left , The nearer position from the speck to the left refracting surface is +0.08m. so this lead me to (1.33/0.08) +1/(v) = (1-0.33)/(-0.12) , my v is -0.05m . r is negative because the refracting surface is concave to the object. i just can't understand why I am wrong ? please help! p/s : the speck is inside the spherical (ball) am i right?
 
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  • #3
Simon Bridge said:
The spec is inside the ball, this is correct.
But you have to account for the light going from a slow to a fast medium (it bends the other way).
Did you try checking your result via a ray diagram?

http://www.math.ubc.ca/~cass/courses/m309-01a/chu/MirrorsLenses/refraction-curved.htm

[edit]

... is that a typo or what you did?

this is what i did exactly ..anything wrong with it?
 
  • #4
Simon Bridge said:
The spec is inside the ball, this is correct.
But you have to account for the light going from a slow to a fast medium (it bends the other way).
Did you try checking your result via a ray diagram?

http://www.math.ubc.ca/~cass/courses/m309-01a/chu/MirrorsLenses/refraction-curved.htm

[edit]

... is that a typo or what you did?

sorry it should be (1.33/0.08) +1/(v) = (1.33-0.33)/(-0.12) where my v is -0.05m . what's wrong with it?
 
  • #5
sorry it should be (1.33/0.08) +1/(v) = (1.33-0.33)/(-0.12) where my v is -0.05m . what's wrong with it?
... so it was a typo - as you wrote it out it was wrong as you noticed.

Did you try verifying your result sing a ray diagram?
Did you check what you did against the link I gave you?
 
  • #6
I checked thru the link. But I stillcant figure out which part I did wrongly.
 
  • #7
Did you try verifying your result using a ray diagram?
Basically your method is correct but you look like you keep making mistakes in substitution.

it should be (1.33/0.08) +1/(v) = (1.33-0.33)/(-0.12)
... which one is n1 and which n2?

I'm getting different values of v to you for the same calculations.
Show me the arithmetic one step at a time.
 
Last edited:
  • #8
Simon Bridge said:
Did you try verifying your result using a ray diagram?
Basically your method is correct but you look like you keep making mistakes in substitution.

... which one is n1 and which n2?

I'm getting different values of v to you for the same calculations.
Show me the arithmetic one step at a time.

n1= 1.33 , n2 = 1.00 , so n1-n2= 1.33-1.00
 
  • #9
desmond iking said:
n1= 1.33 , n2 = 1.00 , so n1-n2= 1.33-1.00
In the equation you are using, is that n1-n2 in the denominator, or n2-n1?
The equation is also at the bottom of that webpage I gave you earlier.
 
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  • #10
Simon Bridge said:
In the equation you are using, is that n1-n2 in the denominator, or n2-n1?
The equation is also at the bottom of that webpage I gave you earlier.


i am using modulus of n2-n1, as in the book. is there's anything wrong with the equation?
 

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  • #11
i am using modulus of n2-n1, as in the book. is there's anything wrong with the equation?
... well, only that you just told me that you did:
it should be (1.33/0.08) +1/(v) = (1.33-0.33)/(-0.12)
...

But you told me that n1=1.33 and n2=1 (post #1) ... so where did that 0.33 come from, and why is n1 first?
What you wrote there does not match your notes, and nor does it match the link I gave you.
surely: n2 - n1 = 1 - 1.33?

Try doing the algebra one line at a time. Be super careful: you appear to have a blind spot there.
 
  • #12
Simon Bridge said:
... well, only that you just told me that you did: ...

But you told me that n1=1.33 and n2=1 (post #1) ... so where did that 0.33 come from, and why is n1 first?
What you wrote there does not match your notes, and nor does it match the link I gave you.
surely: n2 - n1 = 1 - 1.33?

Try doing the algebra one line at a time. Be super careful: you appear to have a blind spot there.

n1 = 1.33 , n2= 1.00 , u = +0.08 , n1-n2 = 1.33-1.00=0.33 , r = -0.12
(1.33/0.08) + (1/v)= (0.33/-0.12) , the v i get is still -0.0516m . i can't figure out which part i did wrongly
 
  • #13
Simon Bridge said:
... well, only that you just told me that you did: ...

But you told me that n1=1.33 and n2=1 (post #1) ... so where did that 0.33 come from, and why is n1 first?
What you wrote there does not match your notes, and nor does it match the link I gave you.
surely: n2 - n1 = 1 - 1.33?

Try doing the algebra one line at a time. Be super careful: you appear to have a blind spot there.

but suprisingly if I take r = +0.12 , i got the ans which is 0.072m . by taking r= -0.12 , i assume the refreacting surface is concave relative to the incident light ray.
 
  • #14
We don't normally do the problem for people, but you are stuck on the substitution and you have sincerely tried hard.
n1 = 1.33 , n2= 1.00 , u = +0.08 , n1-n2 = 1.33-1.00=0.33 , r = -0.12
(1.33/0.08) + (1/v)= (0.33/-0.12)
... the bold face is incorrect.
What is supposed to go in there is n2-n1 and you wrote down n1-n2.
It is very important to write down the equation you are using with all symbols before you do any substituting.

Using:$$n_1=1.33,\; n_2=1,\; o=8\text{cm},\; v=\text{?},\; R=-12\text{cm}$$
$$\frac{n_1}{o}+\frac{n_2}{v}=\frac{n_2-n_1}{R}\\
\implies \frac{1.33}{8}+\frac{1}{v} = \frac{1-1.33}{-12} = \frac{0.33}{12}\\
\implies v= \left( \frac{0.33}{12}- \frac{1.33}{8} \right)^{-1} = -7.21\text{cm}$$ ... now do you see what happened?
 

Related to Geometrical optics- position of image

1. What is the difference between a real and virtual image?

A real image is formed when light rays actually converge at a specific point, while a virtual image is formed when light rays appear to converge at a specific point but do not actually do so. In other words, a real image can be projected onto a screen, while a virtual image cannot.

2. How is the position of an image determined in geometrical optics?

The position of an image is determined by tracing the path of light rays from the object to the mirror or lens, and then using the laws of reflection or refraction to determine where those light rays converge or appear to converge. The point of convergence is the position of the image.

3. Can the position of an image be determined for any type of mirror or lens?

Yes, the position of an image can be determined for any type of mirror or lens as long as the laws of reflection or refraction are followed. However, the specific calculations and equations used may vary depending on the shape and characteristics of the mirror or lens.

4. How does the distance between an object and a mirror or lens affect the position of the image?

The distance between an object and a mirror or lens can affect the position of the image in a few ways. If the distance is greater, the image will be smaller and farther away from the mirror or lens. If the distance is shorter, the image will be larger and closer to the mirror or lens. Additionally, if the object is placed at a certain distance from the mirror or lens known as the focal length, the image will be formed at infinity.

5. Can the size of an image be changed by adjusting the position of the object?

Yes, the size of an image can be changed by adjusting the position of the object. If the object is moved closer to the mirror or lens, the image will become larger. If the object is moved farther away, the image will become smaller. The specific change in size will depend on the distance between the object and the mirror or lens, as well as the focal length and characteristics of the mirror or lens.

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