A) 0.08 m
B) 0.58 m
C) 0.68 m
D) 0.2 m
Correct Answer: D
Solution :
By lens makers formula \[\frac{1}{{{f}_{air}}}={{(}_{a}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?..(i) \[\frac{1}{f\omega }={{(}_{\omega }}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] As \[_{\omega }{{\mu }_{g}}=\frac{_{a}{{\mu }_{g}}}{_{a}{{\mu }_{\omega }}}\] \[\therefore \] \[\frac{1}{f\omega }=\left( \frac{_{a}{{\mu }_{g}}}{_{a}{{\mu }_{\omega }}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?..(ii) Dividing eq. (i) by eq. (ii)/we get \[\frac{{{f}_{\omega }}}{{{f}_{a}}}=\frac{{{{{(}_{a}}{{\mu }_{g}}-1)}_{m}}{{\mu }_{\omega }}}{{{(}_{a}}{{\mu }_{g}}{{-}_{a}}{{\mu }_{\omega }})}\] \[{{f}_{\omega }}=\frac{{{{{(}_{a}}{{\mu }_{g}}-1)}_{a}}{{\mu }_{\omega }}}{{{(}_{a}}{{\mu }_{g}}{{-}_{a}}{{\mu }_{\omega }})}{{f}_{a}}=\frac{(1.5-1)\times 1.33}{(1.5-1.33)}\times 0.2\] so, \[{{f}_{\omega }}-{{f}_{a}}=0.78-0.2=0.58\]
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