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Question
A point object is placed on the principal axis of a convex spherical surface of radius of curvature R, which separates the two media of refractive indices n1 and n2 (n2 > n1). Draw the ray diagram and deduce the relation between the object distance (u), image distance (v) and the radius of curvature (R) for refraction to take place at the convex spherical surface from rarer to denser medium.
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Solution
A refracting surface which forms a part of a sphere of transparent refracting material is called a spherical refracting surface.
The above figure shows the geometry of formation of image I of an object O and the principal axis of a spherical surface with centre of curvature Cand radius of curvature R.
Assumptions:
(i) The aperture of the surface is small compared to other distance involved.
(ii) NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.
`tan ∠NOM = (MN)/(OM)`
`tan ∠NCM =(MN)/(MC) `
`tan ∠NIM = (MN)/(MI)`
For ΔNOC, i is the exterior angle.
∴ i = ∠NOM + ∠NCM
`i = (MN)/(OM)+(MN)/(MC) ........... (1)`
Similarly, r = ∠NCM − ∠NIM
`i.e.,r = (MN)/(MC) - (MN)/(MI) ......(2) `
By Snell’s law,
n1sini = n2sinr
For small angles,
n1i= n2 r
Substituting the values of i and r from equations (i) and (ii), we obtain
`n_1( (MN)/(OM)+(MN)/(MC))= n_2((MN)/(MC) - (MN)/(MI)) `
or,`n_1/(OM) +n_2/(MI)= (n_2 - n_1)/(MC) ........... (3)`
Applying new Cartesian sign conventions,
OM = − u, MI = + v, MC = + R
Substituting these in equation (3), we obtain
`n_2/v-n_1/u = (n_2-n_1)/R ...... (4)`
Above equation is the required equation.
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