Lenses and Lasers: Focal Length of 10 cm Sphere

In summary, the conversation discusses the use of a gypsy's crystal ball as a thick lens with a specific radius of curvature and index of refraction. Two equations, the thick lens equation and the Gaussian formula for spherical surfaces, are presented as possible methods for determining the focal length. The participants also discuss the use of a matrix method and the need to consider assumptions and approximations when using these equations.
  • #1
PH^S!C5
8
0

Homework Statement



A gypsy's crystal ball can act as a thick lens whose centre thickness is twice the radius of curvature of the surface. For a 10 cm sphere of glass with n=1.5 what is the focal length?

Like this? : http://postimage.org/image/niisv7bq9/

Homework Equations



I am not sure if I can simply use the thick lens equation:
1/f = (n-1)*[ (1/r1)-(1/r2)+ ((n-1)*tc)/(n*r1*r2) ]
n=1.5 for glass, and tc = diameter of ball =10cm, since 2tc=r1 then r1=5cm.

or the Gaussian formula for spherical surfaces: f=(n1*r)/(n2-n1) where n1 is the index of refraction on medium of light origin i.e. n1=1 for air; and n2 is the index on medium entered i.e. 1.5 for glass.

The Attempt at a Solution



By the thick lens equation: 1/f= (1.5-1)*[((1.5-1)*10)/(1.5*5*5)]= 0.06666
such that f=15cm.
By the Gaussian formula for spherical surfaces: f=(1*5)/(1.5-1)=10 cm

Question1: which one is correct and how can you determine this?
Question 2: I don't know if r1=-r2 as if one lens were convex (+) and the other concave (-) is taken into account by the formula(s) or if it has to be included at all! Please help! Thanks for your time.
 
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  • #2
One is 15cm and the other 10cm ... that's a big difference!

Question1: which one is correct and how can you determine this?
The way to figure that out is to look at how these equations were derived: what are the assumptions and approximations?

(Note: if desperate, you can always draw a scale diagram and plot a bunch of rays.)

Question 2: I don't know if r1=-r2 as if one lens were convex (+) and the other concave (-) is taken into account by the formula(s) or if it has to be included at all! Please help! Thanks for your time.
Work this out the same way you work out Q1 (above).

Have you tried using the matrix method directly?
 
  • #3
Hi Simon Bridge, thanks for your reply...

I have tried using the following matrix:
[\
\begin{bmatrix}
1 & 0 \\
(n2-n1)/R2 & 1
\end{bmatrix}
x
\begin{bmatrix}
1 & d/n2 \\
0 & 1
\end{bmatrix}
x
\begin{bmatrix}
1 & 0 \\
(n2-n1)/R2 & 1
\end{bmatrix}
=
\begin{bmatrix}
0.934 & 0.67 \\
0.9 & 0.1
\end{bmatrix}\]

The final matrix doesn't make sense though...
 
  • #4
That would be the thick lens method again.
What are the approximations used for this method?
What are the assumptions?

What is it you feel does not make sense?
 
  • #5




I would first like to clarify that the use of the term "gypsy" can be considered derogatory and offensive to the Romani community. I would suggest using a different term or simply referring to it as a crystal ball.

Moving on to the question, the correct formula to use in this scenario would be the thick lens equation. This is because the crystal ball is acting as a thick lens, with a center thickness that is twice the radius of curvature of the surface. The Gaussian formula is used for thin lenses, where the thickness can be neglected.

In this case, the radius of curvature of the surface is equal to half the diameter of the ball (10cm/2 = 5cm). Therefore, the correct formula to use would be:

1/f = (n-1)*[(1/r1) - (1/r2) + ((n-1)*tc)/(n*r1*r2)]

Where n = 1.5 (index of refraction for glass), tc = 10cm (diameter of ball), r1 = 5cm (radius of curvature of surface), and r2 = -5cm (since the surface is concave).

Plugging in these values, we get:

1/f = (0.5)*[(1/5) - (1/-5) + ((0.5)*10)/(1.5*5*-5)]

1/f = 0.1333

Therefore, the focal length of the crystal ball would be 7.5 cm.

To answer your second question, the sign convention for the radius of curvature in the thick lens equation is as follows:

- r1 and r2 are positive for convex surfaces
- r1 and r2 are negative for concave surfaces

In this case, since the surface of the crystal ball is concave, the radius of curvature is negative. It is important to include the sign in the formula to ensure the correct calculation of the focal length.

I hope this explanation helps. Please let me know if you have any further questions.
 

Related to Lenses and Lasers: Focal Length of 10 cm Sphere

1. What is the focal length of a 10 cm sphere?

The focal length of a 10 cm sphere, or the distance between the center of the sphere and the point where light rays converge, is approximately 5 cm.

2. How is the focal length of a lens or sphere calculated?

The focal length of a lens or sphere can be calculated using the formula 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. In the case of a 10 cm sphere, u and v are both equal to 10 cm, resulting in a focal length of approximately 5 cm.

3. What is the relationship between the focal length and the curvature of a lens or sphere?

The focal length of a lens or sphere is inversely proportional to its curvature. This means that a lens or sphere with a smaller curvature will have a longer focal length, while a lens or sphere with a larger curvature will have a shorter focal length.

4. How does the focal length of a 10 cm sphere affect the image formed by it?

The focal length of a 10 cm sphere determines the size and clarity of the image formed. A shorter focal length will result in a smaller, more magnified image, while a longer focal length will result in a larger, less magnified image.

5. Can the focal length of a 10 cm sphere be changed?

The focal length of a 10 cm sphere is a characteristic of the sphere itself and cannot be changed. However, by using different lenses or adjusting the distance between the object and the sphere, the overall focal length of the optical system can be altered.

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