Lens-maker-formula-derivation

Question:I've wasted ages trying to do these problems! please help. An object 50.0 mm high is 39.0 cm away from a concave spherical mirror with a radius of curvature of 46.0 cm. (i) How far from the mirror is the image? (ii) What is the absolute size of the image? (iii) What is the nature of the final image? (b) A picture is behind a layer of glass 2.0 mm thick. The refractive index of the glass is 1.63 . How far behind the front surface of the glass does the picture appear to be? (c) A lens of focal length 43.0 cm is made of glass with index of refraction 1.48 . If one side of the lens is flat, what must the radius of curvature of the other side be?

Answers:a) The image is formed beyond r = 2f in front of the mirror. i) Image distance is v = fu/(u - f) = 23x39/(39 - 23) = 56 cm ii) Magnification = Height of image/height of object = v/u = 56/39 . size of the image = (56/39 )x 5 = 7.18 cm iii) real image. b) n = real depth /apparent depth, apparent depth = real depth / n = 2/1.63 = 1.227m c)The lens maker's formula is 1/f = (n-1)x(1/R1 - 1/R2) Since one surface is plain its radius of curvature is infinite and 1/R = 0 1/f = (n-1)/R R = 0.48/43 = 20.64 cm