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