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Obtain the relative between object distance and image distance in terms of refractive index of medium and radius of curvature for spherically curved surface. |
Answer» Solution :Figure shows the geometry of formation of image I of an object O on the principal axis of a spherical surface with centre of CURVATURE C and RADIUS of curvature R. The rays are INCIDENT from a medium of refractive index n, to another of refractive index `n_2`. NM will be taken to be NEARLY equal to the length of the perpendicular from the point N on the principal axis. We have for small angles, `tan angleNOM=(MN)/(OM) ~~ angle NOM=(MN)/(OM)` `tan angle NCM=(MN)/(MC) ~~ angle NCM =(MN)/(MC)` `tan angle NIM = (MN)/(MI) ~~ angle NIM =(MN)/(MI)` Now, for `triangle`NOC i is the exterior angle. THEREFORE, `i= angle NOM+ angle NCM` `therefore i =(MN)/(OM)+(MN)/(MC)` ... (1) `implies`Similarly, for `triangleNCI,angle NCM=angle NIM+angle INC` `therefore angle NCM=angle NIM+angle INC` `(MN)/(MC)=(MN)/(MI)+r` `therefore r =(MN)/(MC)-(MN)/(MI)` ... (2) Now,by Snell.s law, `n_1sin i=n_2 sin r` Substituting i and r from equation (1) and (2), For small angles sin i `~~` i and sin r `~~` r `therefore n_1i=n_2r` `therefore n_1((MN)/(OM)+(MN)/(MC)) = n_2((MN)/(MC)-(MN)/(MI))` `therefore(n_1)/(OM)+(n_1)/(MC)=(n_2)/(MC)-(n_2)/(MI)` [ `therefore` Dividing by MN] Now taking OM =-u, MC=+R and MI =+v `therefore (n_1)/(-u)+(n_1)/(+R)=(n_2)/(+R)-(n_2)/(+v)` `therefore (n_1)/(-u)+(n_2)/(+v)=(n_2-n_1)/(+R)` or `(n_2)/(v)-(n_1)/(u)=(n_2-n_1)/(R)` If is true for any curved spherical surface. |
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