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Derive the equation for effective focal lengthfor lenses in contact. |
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Answer» Solution :Consider a two lenses of focal length `f_(1) and f_(2)` arrangedcoaxially but separated by a distance d can be considered. For a parallel ray that falls on the arrangement, the two lenses produces deviations `delta_(1) and delta_(2)` respectively and The net deviation `DELTA` is. `delta = delta_(1) + delta_(2)` From Angle of deviation in lens equation, `delta = (h)/(F)` `delta_(1) = (h_(1))/(f_(1)):delta_(2) = (h_(2))/(f_(2)) and delta = (h_(1))/(f)` The equation (1) BECOMES, `(h_(1))/(f) = (h_(1))/(f_(1)) + (h_(2))/(f_(2))` From the geometry. `h_(2)-h_(1)=P_(2)G-P_(2)C=CG` `h_(2) - h_(1) = BG tan delta_(1) = BG delta_(1)` `h_(2) - h_(1) = d(h_(1))/(f_(1))` `h_(2) = h_(1) + d(h_(1))/(f_(1))` Substituting the above equaiton in Equation (3) `(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_(1)f_(2))` The above equation COULD be used to find the EQUIVALENT focal lenght. 
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