A) \[{{f}_{g}}>{{f}_{r}}\]
B) \[{{f}_{v}}<{{f}_{r}}\]
C) \[{{f}_{v}}>{{f}_{r}}\]
D) \[{{f}_{v}}={{f}_{r}}\]
Correct Answer: B
Solution :
Key Idea: The conclusion can be drawn from Lens makers formula and Cauchys relation. Lens makers formula is \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] or \[\frac{1}{f}\propto \mu \] or \[f\propto \frac{1}{\mu }\] ??.(i) According to Cauchys formula \[\mu \propto \frac{1}{\lambda }\] ?..(ii) From Eqs. (i) and (ii), we get \[f\propto \lambda \] Hence, focal length of a converging lens is maximum for red colour (highest wavelength) and minimum for violet colour (lowest wavelength) i.e., \[{{f}_{v}}<{{f}_{r}}\] NOTE: Lens makers formula is so called because it can be used to determine the values of \[{{R}_{1}}\] and \[{{R}_{2}}\] that are needed for a given refractive index and a desired focal length f.
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