A) the focus shifts to infinity
B) the focal point shifts towards the lens by a small distance
C) the focal point shifts away from die lens by a small distance
D) the focus remains undisturbed
Correct Answer: A
Solution :
The combined focal length of plano-convex lens \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] Here :\[{{f}_{1}}=\infty \] (for plane surface), \[{{f}_{2}}=f\](say) \[\therefore \] \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] \[\Rightarrow \] \[F=f\] Now, when concave lens of same focal length is joined to first lens, then combined focal length \[\frac{1}{F}=\frac{1}{{{F}_{1}}}+\frac{1}{{{F}_{2}}}\] \[=\frac{1}{f}-\frac{1}{f}\] \[(\because \,{{F}_{1}}=f,\,{{F}_{2}}=-f)\] = 0 or \[F=\infty \] Thus, the image can be foccussed on infinity \[(\infty )\] or focus shifts to infinity.
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