Finding the Image Distance for a Mirrored Sphere

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AI Thread Summary
To find the image distance for a mirrored hemisphere, the problem involves a point object placed at a distance from a sphere of radius R, which is then halved with the planar surface silvered. The relevant formula for image formation is applied across three optical surfaces: the spherical surface, the planar surface, and the mirror. The user attempts to calculate the image distances sequentially but struggles to arrive at the correct answer. There is confusion over the application of the formula and the treatment of the surfaces involved. Ultimately, a clear understanding of the optical principles and careful application of the equations is necessary to solve the problem.
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Homework Statement


There is a sphere of Radius R. A point object is kept at a distance b from it.Now the sphere is halfed about an axis perpendicular line joining the object and centre of the sphere.The planar surface of the hemisphere away from the object is silvered.Find the distance of the image from the sphere?

Homework Equations


\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}
where symbols have their usual meanings
also silvering means making it a mirror

The Attempt at a Solution


Now there are basically three optical surfaces
the first spherical surface
then the planar and then the mirror.
What i did was to apply the formula for 1st surface then find the image which acts as object for second interface then use the found out image as object for mirror .then find the third image(i.e due to mirror) then this acts as an object again for the planar refracting surface .
then agian find the image and finally use it as an object for the curved surface .But still i am unable to get the answer :confused:
 
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pardesi said:
Now there are basically three optical surfaces
the first spherical surface
then the planar and then the mirror.
I would treat this as two surfaces: The first spherical surface, then the plane mirror.
 
ok let me have a go at that
but why is that so
 
no that's not working
 
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