Solution :
a). The focal length of the lens is given as :
![$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7Bf%7D%3D%28n-1%29%5Cleft%28%5Cfrac%7B1%7D%7BR_1%7D-%5Cfrac%7B1%7D%7BR_2%7D%5Cright%29%24)
![$\frac{1}{f}=(1.70-1)\left(\frac{1}{\infty}-\frac{1}{-13}\right)$](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7Bf%7D%3D%281.70-1%29%5Cleft%28%5Cfrac%7B1%7D%7B%5Cinfty%7D-%5Cfrac%7B1%7D%7B-13%7D%5Cright%29%24)
![$f=18.57 \ cm$](https://tex.z-dn.net/?f=%24f%3D18.57%20%5C%20cm%24)
The thin lens equation is :
![$\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7Bd_i%7D%2B%5Cfrac%7B1%7D%7Bd_o%7D%24)
The distance of the image from the lens is
![$d_i=\left(\frac{1}{f}-\frac{1}{d_o}\right)^{-1}$](https://tex.z-dn.net/?f=%24d_i%3D%5Cleft%28%5Cfrac%7B1%7D%7Bf%7D-%5Cfrac%7B1%7D%7Bd_o%7D%5Cright%29%5E%7B-1%7D%24)
![$d_i=\left(\frac{1}{18.57}-\frac{1}{22.5}\right)^{-1}$](https://tex.z-dn.net/?f=%24d_i%3D%5Cleft%28%5Cfrac%7B1%7D%7B18.57%7D-%5Cfrac%7B1%7D%7B22.5%7D%5Cright%29%5E%7B-1%7D%24)
![$=1.06 \ m$](https://tex.z-dn.net/?f=%24%3D1.06%20%5C%20m%24)
The magnification of the lens is :
![$m=-\frac{d_i}{d_o}=\frac{h_i}{h_o}$](https://tex.z-dn.net/?f=%24m%3D-%5Cfrac%7Bd_i%7D%7Bd_o%7D%3D%5Cfrac%7Bh_i%7D%7Bh_o%7D%24)
The height of the image is :
![$h_i=-\frac{d_i}{d_o}(h_o)$](https://tex.z-dn.net/?f=%24h_i%3D-%5Cfrac%7Bd_i%7D%7Bd_o%7D%28h_o%29%24)
![$h_i=-\frac{106.3 \ cm}{22.5 \ cm}(3.75 \ mm)$](https://tex.z-dn.net/?f=%24h_i%3D-%5Cfrac%7B106.3%20%5C%20cm%7D%7B22.5%20%5C%20cm%7D%283.75%20%5C%20mm%29%24)
![$=-17.7 \ cm$](https://tex.z-dn.net/?f=%24%3D-17.7%20%5C%20cm%24)
Thus, the real and the inverted image is formed at a distance of 1.06 m. And the size of the image is 17.7 cm
b). When the lens is reversed.
The focal length of the lens is given as :
![$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7Bf%7D%3D%28n-1%29%5Cleft%28%5Cfrac%7B1%7D%7BR_1%7D-%5Cfrac%7B1%7D%7BR_2%7D%5Cright%29%24)
![$\frac{1}{f}=(1.70-1)\left(\frac{1}{13}-\frac{1}{\infty}\right)$](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7Bf%7D%3D%281.70-1%29%5Cleft%28%5Cfrac%7B1%7D%7B13%7D-%5Cfrac%7B1%7D%7B%5Cinfty%7D%5Cright%29%24)
![$f=18.57 \ cm$](https://tex.z-dn.net/?f=%24f%3D18.57%20%5C%20cm%24)
The thin lens equation is :
![$\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7Bd_i%7D%2B%5Cfrac%7B1%7D%7Bd_o%7D%24)
The distance of the image from the lens is
![$d_i=\left(\frac{1}{f}-\frac{1}{d_o}\right)^{-1}$](https://tex.z-dn.net/?f=%24d_i%3D%5Cleft%28%5Cfrac%7B1%7D%7Bf%7D-%5Cfrac%7B1%7D%7Bd_o%7D%5Cright%29%5E%7B-1%7D%24)
![$d_i=\left(\frac{1}{18.57}-\frac{1}{22.5}\right)^{-1}$](https://tex.z-dn.net/?f=%24d_i%3D%5Cleft%28%5Cfrac%7B1%7D%7B18.57%7D-%5Cfrac%7B1%7D%7B22.5%7D%5Cright%29%5E%7B-1%7D%24)
![$=1.06 \ m$](https://tex.z-dn.net/?f=%24%3D1.06%20%5C%20m%24)
The magnification of the lens is :
![$m=-\frac{d_i}{d_o}=\frac{h_i}{h_o}$](https://tex.z-dn.net/?f=%24m%3D-%5Cfrac%7Bd_i%7D%7Bd_o%7D%3D%5Cfrac%7Bh_i%7D%7Bh_o%7D%24)
The height of the image is :
![$h_i=-\frac{d_i}{d_o}(h_o)$](https://tex.z-dn.net/?f=%24h_i%3D-%5Cfrac%7Bd_i%7D%7Bd_o%7D%28h_o%29%24)
![$h_i=-\frac{106.3 \ cm}{22.5 \ cm}(3.75 \ mm)$](https://tex.z-dn.net/?f=%24h_i%3D-%5Cfrac%7B106.3%20%5C%20cm%7D%7B22.5%20%5C%20cm%7D%283.75%20%5C%20mm%29%24)
![$=-17.7 \ cm$](https://tex.z-dn.net/?f=%24%3D-17.7%20%5C%20cm%24)
Thus, when the lens is reversed, a real and an inverted image is formed at a distance of 1.06 m. And the size of the image is 17.7 cm