Optics Problem: What am I doing wrong?

[xenogizmo]

Hey Everyone..
I'm having trouble with this question, I think I have this question figured out but my answer is coming out incorrect... Tell me if there is a flaw in my logic.. question is:

"Suppose that a luminous star of radius R1=1.74×108 m is surrounded by a uniform atmosphere extending up to a radius R2=4.28×108 m and with index of refraction n=1.82. When the sphere is viewed from a location far away in vacuum, what is its apparent radius?"

What I did was consider the outer edge of the atmosphere a spherical refracting surface.. and using the equation: n1/P + n2/q = (n2-n1)/R
(sorry for not using latex)

And then I took a point from the star's surface, and take the distance form the atmosphere to that point the image distance (where P is just R2 - R1).
I also took n1 as 1.82 and n2 as 1 since it's just vacuum.. and according to the sign conventions the radius should be negative..

After I got the image distance (which was negative, a virtual image), I subtracted it's absolute value from R2, and that should be apparent radius. Am I wrong?

Thx,
Xeno

[xenogizmo]

btw, the answer I got when I did this was 2.313x10^8 m

there's a hint on that question but it made sense to me.. Here it is just in case: "What is special about the case R2 > n*R1? "

[xenogizmo]

Ideas anyone?

[GCT]

[quote]Suppose that a luminous star of radius R1=1.74×108 m is surrounded by a uniform atmosphere extending up to a radius R2=4.28×108 m and with index of refraction n=1.82. When the sphere is viewed from a location far away in vacuum, what is its apparent radius?"

What I did was consider the outer edge of the atmosphere a spherical refracting surface.. and using the equation: n1/P + n2/q = (n2-n1)/R
(sorry for not using latex)[\quote]

I got a different answer (3.18 x 10^8) using the same method as yours. Although this might not be correct, retry the problem precisely.

[ehild]

[quote]Suppose that a luminous star of radius R1=1.74×108 m is surrounded by a uniform atmosphere extending up to a radius R2=4.28×108 m and with index of refraction n=1.82. When the sphere is viewed from a location far away in vacuum, what is its apparent radius?"

[\quote]

I got a different answer (3.18 x 10^8)

This is an other approach, with the same result as yours. See attached picture.

Consider the ray emerging from a surface point S of the star. It is refracted at the edge of the atmosphere, and travels toward the observer. He sees that point at the apparent distance d from the center of the star.

d=R_2 \sin {\alpha}

We have the relation

\frac{sin{\beta}}{\sin{\gamma}}=\frac{R_1}{R_2}

between the angles of the yellow triangle.

According to Snell's law,

n \sin{\beta} = \sin{\alpha}

These result in

\sin{\alpha} = \frac{R_1}{R_2} n \sin {\gamma}

The apparent radius of the star is obtained with the maximum of sin(alpha).

R_{ap} = R_2 (\sin{\alpha})_{max}

This can be R2 if nR1 is greater than or equal to R2. Otherwise it is nR1, when gamma = pi/2.

nR1 = 1.74×108*1.82 = 3.17×108 m. This is less than R2, so Rap = 3.17×108 m.

ehild

[GCT]

I found the original problem online and it seems that they wanted the answer in your form, since one had to incorporate the given R2>nR1