Answer:
a) 5.39 m
b) 0.18 m^-1
Explanation:
The focal length of this kind of optical system is given by:
[tex]\frac{1}{f}\approx(n-1)[\frac{1}{R_1}-\frac{1}{R_2}][/tex]
you replace the values of the parameters in order to find the focal length:
[tex]\frac{1}{f}=(1.59-1)[\frac{1}{1.15m}-\frac{1}{1.80m}]=0.18m^-1\\\\f=\frac{1}{0.18}m=5.39m[/tex]
hence, the magnitude of the focal length is 5.39m
The power of the lens is:
[tex]P=\frac{1}{f}=0.18\ m^{-1}[/tex]
the power of the lens is 0.18 m^-1
