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))`
