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The image of a white object in with light formed by a lens is usually colored and blurred. This defect of image is called chromatic aberration and arises due to the fact that focal length of a lens is different for different colours. As R .I. mu of lens is maximum for violet while minimum for red, violet is focused nearest to the lenswhile red farthest from it as shown in figure. As a result of this, in case of convergent lens if a screen is placed at F_(v) center of the image will be violet and focused while sides are red and blurred. While at F_(R) , reverse is the case, i.e ., center will be red and focused while sides violet and blurred. The differece between f_(v) and f_(R) is a measure of the longitudinal chromatic aberration (L.C.A),i.e., L.C.A.=f_(R)-f_(v)=-df with df=f_(v)-f_(R)...........(1) However, as for a single lens, (1)/(f)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))].............(2) rArr -(df)/(f^(2))=dmu[(1)/(R_(1))-(1)/(R_(2))]...............(3) Dividing E1n. (3) by (2) : -(df)/(f)=(dmu)/((mu-1))=omega,[omega=(dmu)/((mu-1))] "dispersive power",.........(4) And hence, from Eqns. (1) and (4) , L.C.A.=-df=omegaf Now, as for a single lens neither f nor omega zero, we cannot have a single lens free from chromatic aberration. Condition of Achromatism : In case of two thin lenses in contact (1)/(F)=(1)/(F_(1))+(1)/(F_(2))i.c,. -(dF)/(F^(2))=(df_(1))/(f_(1)^(2))-(df_(2))/(f_(2)^(2)) The combination will be free from chromatic aberration if dF=0 i.e., (df_(1))/(f_(1)^(2))+(df_(2))/(f_(2)^(2))=0 which with the help of Eqn. (4) reduces to (omega_(1)f_(1))/(f_(1)^(2))+(omega_(2)f_(2))/(f_(2)^(2))=0 ,i.e., (omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0........(5) This condition is called condition of achromatism (for two thin lenses in contact ) and the lens combination which satisfies this condition is called achromatic lems, from this condition, i.e., form Eqn. (5) it is clear the in case of achromatic doublet : Since, if omega_(1)=omega_(2), (1)/(f_(1))+(1)/(f_(2))=0i.e., (1)/(F)=0 or F=infty i.e., combination will not behave as a lens, but as a plane glass plate. (2) As omega_(1) and omega_(2) are positive quantities, for equation (5) to hold, f_(1) and f_(2) must be of opposite nature, i.e., if one of the lenses is converging the other must be diverging. (3) If the achromatic combination is convergent, f_(C)ltf_(D) and as (f_(C))/(f_(d))=(omega_(C))/(omega_(D)), omega_(C)ltomega_(d) i.e., in a convergent achromatic doublet, convex lens has lesses focal legth and dispersive power than the divergent one. Chromatic aberration of a lens can be corrected by :

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PROVIDING DIFFERENT suitable curvatures of its two surfaces.
proper POLISHING of its two surfaces.
suitably conbining it with ANOTHER lens.
reducing its aperture.

Solution :N//a


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