A) 10.5
B) 12
C) 13.5
D) 15.5
Correct Answer: C
Solution :
Ramsdens eye-piece consists of two co-axial plano-convex lenses of focal length \[f\] and \[{{f}_{1}}\] separated by a distance \[\frac{2f}{3},\] \[\frac{2f}{3}=12\] \[\Rightarrow \] \[f=\frac{12\times 3}{2}=18\,cm\] So, equivalent focal length \[\frac{1}{F}=\frac{1}{f}+\frac{1}{f}-\frac{d}{{{f}^{2}}}\] \[=\frac{1}{18}+\frac{1}{18}-\frac{12}{{{(18)}^{2}}}\] \[=\frac{1}{9}-\frac{1}{27}\] \[=\frac{2}{27}\] \[\therefore \] \[f=\frac{27}{2}=13.5\,cm\]
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