Write the relation between `B_(E), B_(H) "and" B_(V)` along with an appropriate diagram .

Explanation

Figure shows the formation of image I of an object O on the principal axis of a convex surface with centre of curvature C.
As the aperture is small the length of NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.
For small angles (assuming paraxiat rays) we have,
`tan (|underline (NOM)) ~~ |underline (NOM) = (MN)/(OM) `
`tan (|underline (NCM)) ~~ |underline (NCM) = (MN)/(MC) `
`tan (|underline (NIM)) ~~ |underline (NIM) = (MN)/(MI) `
Here i =
`|underline(NOM) + |underline(NCM) = (MN)/(OM) + (MN)/(MC)`
and r =
`|underline(NCM) - |underline(NIM) = (MN)/(MC) - (MN)/(MI)`
For small angles, Snell.s law, `n_(1) sini = n_(2)= sinr ` can be written as,
`n_(1) i = n_(2)= r`
`n_(1) ((MN)/(OM) + (MN)/(MC)) = n_(2) ((MN)/(MC) - (MN)/(MI))`
`n_(1) MN((1)/(OM) + (1)/(MC)) = n_(2) MN((1)/(MC) - (1)/(MI))`
`(n_(1))/(OM) + (n_(1))/(MC) = (n_(2))/(MC) - (n_(2))/(MI)`
`(n_(1))/(OM) + (n_(2))/(MI) = (n_(2))/(MC) - (n_(1))/(MC)`
`(n_(1))/(OM) + (n_(2))/(MI) = (n_(2) - n_(1))/(MC)`
Applying the Cartesian sign convention, `OM = - u, MI = + v, MC = + R`
`(n_(1))/(-u) + (n_(2))/(v) = (n_(2) - n_(1))/(R)`
`(n_(2))/(v) - (n_(1))/(u) = (n_(2) - n_(1))/(R)`