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```				   One half of a sphere of radius R & refractive index μ is silvered. A point object is placed in front of the unsilvered portion of the sphere at a distance R from the surface of the sphere. Where will the final image be formed?
```

6 years ago

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```										  in this question
u= distance of object from pole i.e 3R
v = distance of image
f= distance of focal point/
all these distances are taken from pole
therefore a.
u = 3R & R = radius of curvature ; so f= R/2 & R=2f. put this value of R in  u
so u = 6f
thus according to mirror formula
1/v + 1/u = 1/f : put values of u in this equation

so 1/v = 1/f - 1/6f
i.e v= 6f/5 & 6f can be written as 3R
so value of v= 3R/5 : hence at this position image will be formed at this distance from pole .
```
6 years ago

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