Wow you make things so much more clear, I really appreciate your time. I think I finally have it:
For the third surface, the image distance is 64.6+(2*3.4)=71.4 to left of third surface.
1/71.4+1.5/s'=(1.5-1)/-3.4
s'=-9.313 or 9.313 cm to the left of third surface.
For the fourth surface...
Im sorry, stupid mistake by me. So using -11.4:
1.5/-11.4+1/s'=-0.5/3.4
s'=-64.6 cm or 64.6 cm to the left of the second surface.
I read in my textbook that Radius of Curvature is negative when in front of the surface, but was confused with what they meant by in front of the surface. Is...
Oh I see. The distance of that image to the second surface (glass to air on left side of sphere) is -8 cm.
n1/s+n2/s'=(n2-n1)/R
1.5/-8+1/s'=(1-1.5)/3.4
s'=24.7cm or 24.7 cm to the right of the second surface.
If this is correct, continuing onto the third surface (air to glass on...
Using this as my object for the right side of the sphere, i get s=12-8=4cm to right of right side of sphere.
n1/s+n2/s'=(n2-n1)/R
1/4+1.5/s'=(1.5-1)/4
s'=-12 cm, or 12 cm to the left of the right side of the sphere. So the final image is 4 cm to the left of the left side of the sphere...
I read the section on spherical refracting surfaces, and the equation for object-image relationship is:
n1/s+n2/s'=(n2-n1)/R
For this question, the object is at infinity and n1 is air so the equation reduces to
n2/s'=(n2-1)/R
1.5/s'=0.5/4
s'=12 cm to the right of left edge of...
The question is from the Mastering Physics program, the textbook is University Physics 12th Edition (Young and Freedman).
Would i use 1/f=(n-1)(1/R1-1/R2+(n-1)d/nR1R2) ?
Oh ok, so the focal length of the second 'lens' will be -45.3 cm as well. Using the image from the first lens as the object:
1/f=1/s+1/s'
s=45.3cm + Outer radius + inner radius
s=45.3+4+3.4
s=52.7 cm
1/-45.3=1/52.7+1/s'
s'=-24.4
Final image is -24.4+7.4 cm = 17 cm to left of...
As the light continues along its path it reaches the right side of the sphere. Using the same equation:
1/f=(1.5-1)(1/3.4-1/4.0)
f=+45.3 cm
So there will be an image on either side of the glass sphere, 45.3 cm away. The question asks where will the image be formed to the left from the...
When the object is at an infinite distance from the lens, the image distance equals the focal length.
I have two equations now, f=R/2, and 1/f=(n-1)(1/R1-1/R2) (lens makers' equation).
Using the lens makers equation:
1/f=(1.50-1)(1/4.0cm-1/3.4cm)
f=-45.3 cm
Using the first equation...
Homework Statement
A thick-walled wine goblet sitting on a table can be considered to be a hollow glass sphere with an outer radius of 4.00 cm and an inner radius of 3.40 cm. The index of refraction of the goblet glass is 1.50.
A) A beam of parallel light rays enters the side of the empty...
I think I understand: sin(ThetaCritical)=n2/n1
sin(58.7)=1.33/n1
n1=1.5565
Then:
n1=c/v
1.5565=3*10^8/v
v=1.927*10^8 m/s
Thank you for directing me to Total internal reflection, does the rest look correct?
Homework Statement
A large cube of glass has a metal reflector on one face and water on an adjoining face (the figure). A light beam strikes the reflector, as shown. You observe that as you gradually increase the angle of the light beam, if Theta is greater than 58.7 no light enters the water...
Homework Statement
You have been assigned to design brass pistons to slide inside steel cylinders. The engines in which these pistons will be used will operate between 20.0 degrees Celsius and 150.0 degrees Celsius. Assume that the coefficients of expansion are constant over this temperature...