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Draw a ray diagram to show the formation of the real image of a point object due to a convex spherical refracting surface, when the ray of light is travelling from a rarer medium of refractive index nx to a denser medium of refractive index n_(2) . Using this diagram derive the relation between object distance (w), image distance (v), radius of curvature (R) of a convex spherical surface. State the sign convention and the assumptions used. |
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Answer» Solution :Sign Convention: (i) All distances are measured from the pole of the spherical refracting surface. (ii) Distances measured along the direction of incidence of light are taken as positive and the distances measured in a direction opposite to the direction of incidence of light are taken as negative. (iii) Heights upward from principal axis are taken as positive and distances downward from principal axis are taken as negative. Assumptions : (i) The object is a point object situated on the principal axis of the spherical refracting surface. Moreover, the object is situated such that the incident RAY travels from left to right. (ii) Aperture of the refracting surface is small so that values of angle of incidence and angle of refraction are small enough. (iii) A refracting surface is considered as concave or convex as seen from the side of rarer medium only. Refraction formula : Consider a spherical surface XPY, convex towards the rarer medium. As shown in Fig. 9.89 for a point object situated in rarer medium of refractive index n-y, the image is formed in denser medium of refractive index `n_2` at point I. If i and r be the angles of incidence and refraction RESPECTIVELY, then from Snell.s law  `(sin i)/(sin r) = n_(2)/n_(1)` or `n_(1)sini = n_(2)sin r` and for small aperture angles i and r are small and therefore sin i - i and sin r, hence `n_(1)i = n_(2)r` From `triangleOAC`, i `=(alpha + gamma)` and from `triangleIAC, lambda = BETA + r` or `r = lambda -beta`..... (ii) `therefore n_(2)i = n_(2)r` `alpha = tan alpha + gamma` and from `triangleIAC, gamma = beta + r` or `r = gamma - beta` `alpha = tan alpha = (AN)/(ON) = (AN)/(OP), beta = tan beta = (AN)/(IN) = (AN)/(PI)`, and `gamma = tan gamma = (AN)/(NC) =(AN)/(PC)` Substituting these values of `alpha, beta` and `gamm` in (iii), we have `n_(1)[(AN)/(OP) + (AN)/(PC)] =n_(2) [(AC)/(PC) -(AN)/(PI)]` or `n_(1)/(OP) + n_(1)/(PC) = n_(2)/(PC) -n_(2)/(PI)` As per sign convention followed, `OP=-u, PC =+R` and Pl `=+v`. Hence, equation (iii) Becomes `n_(1)/(-u) + n_(2)/(+R) = n_(2)/(+R) -n_(2)/(+v)` which is the relation between object distance u, image distance v and the radius R of the GIVEN spherical surface. |
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