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Derive the equation for effective focal length for lenses in out of contact. |
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Answer» Solution :When two thin lenses are separated by a distance d. (i) Let O be a point object on the principal axis of a lens. OA is the incident ray on the lens at a point A at a height h above the optical centre. (II) The ray is deviated through an angle `delta` and forms the image at I on the principal axis. (iii) The incident and refracted rays subtend the angles, `angleAOP = alpha` and `angleAIP = beta` with the principal axis respectively. In the TRIANGLE `triangleOAI` , the angle of deviation `delta` can be written as, ` delta = alpha + beta "" ....(1)` If the height h is small as compared to PO and Pl the angles `alpha , beta ` and `delta`are also small. Then, ` alpha ~~ tan alpha = (PA)/(PO) , " and " beta ~~ tan beta = (PA)/(PI) "" ....(2)` Then,` delta= (PA)/(PO) + (PA)/(PI) "" ....(3)` Here, `PA = h,PO = - u " and " PI = v` ` delta = (h)/( - u) + h/v = h ((1)/(-u) + (1)/(v)) "" ....(4)` After rearranging ` delta = h (1/v - 1/u) = h/f` ` delta = h/f "" ...(5)` (iv) The above equation tells that the angle of deviation is the ratio of height to the focal length. Now, the case of two lenses of focal length `f_1` and `f_2` arranged coaxially but separated by a distance d can be considered as shown in the below Figure.  (v)For a parallel ray that falls on the arrangement, the two lenses produce deviations `delta_1` and `delta_2` respectively and The net deviation `delta` is. ` delta= delta_1 + delta_2 "" ....(6)` From Equation (5), ` delta_1 = (h_1)/(f_1) , delta_2 = (h_2)/(f_2) " and " delta = (h_1)/(f) "" ....(7)` The equation (6) becomes, ` (h_1)/(f) + (h_1)/(f_1) + (h_2)/(f_2) "" ....(8)` From the geometry, ` h_2 - h_1 = P_2 G - P_2 C= CG` `h_2- h_1 = BG tan delta_1 ~~ BGdelta_1` ` h_2 - h_1 = d(h_1)/(f_1)` ` h_2 = h_1 d(h_1)/(f_1) "" ....(9)` Substituting the above equation in Equation (8). ` (h_1)/(f) = (h_1)/(f_1) + (h_1)/(f_2) + (h_1 d)/(f_1 f_2)` On further simplification, `1/f = (1)/(f_1) + (1)/(f_2) + (d)/(f_1f_2) "" ....(10)` (vi) The above equation could be used to FIND the equivalent focal length. To find the position of the equivalent lens, we can further write from the geometry, ` PP_2 = EG = (GC)/(tan delta)` `PP_2 = EG = (GC)/(tan delta) = (h_1 - h_2)/(tan delta) = (h_1 - h_2)/(delta)` From equations (7) and (9) ` h_2 - h_1 = d (h_1)/(f_1) " and " delta= (h_1)/(f)` `PP_2 = (d (h_1)/(f_1)) xx ((f)/(h_1))` ` PP_2 = (d (f)/(f_1))` |
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