A) +15 cm and - 20 cm
B) - 15 cm and -20 cm
C) +20 cm and -25 cm
D) -20 cm and +25 cm
Correct Answer: A
Solution :
Let focal length of the convex and concave lenses be \[{{f}_{1}}\] and \[{{f}_{2}}\]. In order that this is achromatic combination of lenses, we have \[\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}=\frac{1}{F}\] Give, \[F=60\,cm\] \[\therefore \] \[\frac{1}{60}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] ?(i) Also, condition for lens combination to be achromatic, we have \[\frac{{{\omega }_{1}}}{{{f}_{1}}}+\frac{{{\omega }_{2}}}{{{f}_{2}}}=0\] \[\Rightarrow \] \[\frac{{{\omega }_{1}}}{{{\omega }_{2}}}=-\frac{{{f}_{1}}}{{{f}_{2}}}\]where, \[{{\omega }_{1}}\]and \[{{\omega }_{2}}\]are dispersive powers. we have given, \[\frac{{{\omega }_{1}}}{{{\omega }_{2}}}=\frac{3}{4}\] \[\therefore \] \[\frac{3}{4}=-\frac{{{f}_{1}}}{{{f}_{2}}}\] \[\Rightarrow \] \[{{f}_{2}}=-\frac{4}{3}{{f}_{1}}\] ?(ii) Putting this value in (i), we have \[\frac{1}{60}=\frac{1}{{{f}_{1}}}-\frac{3}{4{{f}_{1}}}\] \[\Rightarrow \]\[{{f}_{1}}=+15\,cm\] \[\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}=\frac{1}{F}\] Give, \[F=60\,cm\] \[\therefore \] \[\frac{1}{60}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] ?(i) Also, condition for lens combination to be achromatic, we have \[\frac{{{\omega }_{1}}}{{{f}_{1}}}+\frac{{{\omega }_{2}}}{{{f}_{2}}}=0\] \[\Rightarrow \] \[\frac{{{\omega }_{1}}}{{{\omega }_{2}}}=-\frac{{{f}_{1}}}{{{f}_{2}}}\]where, \[{{\omega }_{1}}\]and \[{{\omega }_{2}}\]are dispersive powers. we have given, \[\frac{{{\omega }_{1}}}{{{\omega }_{2}}}=\frac{3}{4}\] \[\therefore \] \[\frac{3}{4}=-\frac{{{f}_{1}}}{{{f}_{2}}}\] \[\Rightarrow \] \[{{f}_{2}}=-\frac{4}{3}{{f}_{1}}\] ?(ii) Putting this value in (i), we have \[\frac{1}{60}=\frac{1}{{{f}_{1}}}-\frac{3}{4{{f}_{1}}}\] \[\Rightarrow \]\[{{f}_{1}}=+15\,cm\] \[{{f}_{2}}=-20\,cm\] \[\Rightarrow \] \[{{f}_{2}}=-\frac{4}{3}{{f}_{1}}\] Putting this value in (i), we have \[\frac{1}{60}=\frac{1}{{{f}_{1}}}-\frac{3}{4{{f}_{1}}}\] Hence, one is a convex lens and the other is a concave lens.
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