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?

  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 .