A) \[{{\lambda }_{R}}\le {{\lambda }_{V}}\]
B) \[{{\mu }_{R}}>{{\mu }_{V}}\]
C) \[{{\lambda }_{R}}={{\lambda }_{V}}\]
D) \[{{\mu }_{R}}<\mu {{\,}_{V}}\]
Correct Answer: D
Solution :
The relation between wavelength \[(\lambda )\] of light in an optically dense medium to its wavelength in vacuum is given by \[\mu =\frac{c}{v}\] \[\mu =\frac{f\,\lambda }{f\,\lambda }\] \[\Rightarrow \] \[\mu =\frac{\lambda }{\lambda }\] We know red colour has a larger wavelength (\[{{\lambda }_{R}}=650\] nm) than violet (\[{{}_{V}}=400\]nm), hence \[{{\mu }_{R}}<{{\mu }_{V}}\] Alternative : Focal length \[\frac{1}{f}=(\mu -1)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[\Rightarrow \] \[\frac{1}{f}\propto \mu \,\,\,\,\,\Rightarrow \,\,\,\,\,f\propto \frac{1}{\mu }\] Since, focal length of violet light is less than that of red light, so \[{{\mu }_{R}}<{{\mu }_{V}}\]
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