A plano-convex lens is made of material of refractive index n When a small object is placed 30cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens. Another faint image is observed at a distance of 10cm away from the lens. Which of the following statement(s) is (are) true?

For refraction by the plano-convex lens, object distance is $u=-30cm.$ The size of the image is double the size of the object i.e., magnification $m=\upsilon /u=\pm 2,$ where positive sign is for erect image and negative sign is for inverted image. Thus, the image distance is $\upsilon =60\text{\hspace{0.17em}cm}$ (inverted image) or  $\upsilon =-60\text{\hspace{0.17em}cm}$ (erect image). Apply lens formula,  $\frac{1}{f}=\frac{1}{\upsilon }-\frac{1}{u},$ to get  $f=20\text{\hspace{0.17em}cm}$ (inverted image) and $f=60\text{\hspace{0.17em}cm}$ (erect image).
    
Let R be the radius of the convex surface of the plano-convex lens which is made of material of refractive index n Apply lens-maker’s formula to get its focal length f as
$\frac{1}{f}=(n-1)[\frac{1}{R}-\frac{1}{\infty }]=\frac{n-1}{R}.$             (1)
   
The faint image is formed due to reflection form the convex surface, which acts as a convex mirror. The focal length of this mirror is $f_m=R/2,$ object distance is  $u=-30cm$ and image distance is  $\upsilon =10cm$ (the image by convex mirror is erect, virtual, and is formed towards the right of the pole). Apply the mirror formula, $\frac{1}{f_m}=\frac{1}{\upsilon }+\frac{1}{u},$ to get $f_m=15cm.$ thus, the radius of curvature of the convex surface is  $R=2f_m=30cm.$
 Substitute  $R=30$cm in equation (1) to get  $n=2.5$ for  $f=20$cm (inverted image) and  $n=1.5$ for  $f=60$cm (erect image).