A) increases by a factor of 1.25
B) increases by a factor of 2.5
C) increases by a factor of 1.2
D) decreases by a factor of 1.2
Correct Answer: B
Solution :
Lens-maker?s formula is given by \[\frac{1}{f}=\left( _{a}{{\mu }_{g}}-1 \right)\,\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?. (i) where\[_{a}{{\mu }_{g}}\]. is the refractive index of glass w.r.t. air, \[{{R}_{1}}\,and\,\,{{R}_{2}}\] are the radii of curvature of two surfaces of lens and f is the focal length of the lens. If the lens is immersed in a liquid of refractive index\[{{\mu }_{l}}\], then Here, \[_{l}{{\mu }_{g}}\] is the refractive index of glass w.r.t. liquid. Dividing Eq. (i) by Eq. (ii), we have \[\frac{{{f}_{1}}}{f}=\frac{\left( _{a}{{\mu }_{g}}-1 \right)}{\left( _{l}{{\mu }_{g}}-1 \right)}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\frac{{{f}_{1}}}{f}=\left( \frac{1.5-1}{\frac{1.5}{1.25}-1} \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\frac{{{f}_{1}}}{f}=\frac{0.5\times 1.25}{0.25}=2.5\] Hence, focal length increases by a factor of 2.5.
You need to login to perform this action.
You will be redirected in
3 sec
