A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. Calculate the magnification

Explanation

longitudinal magnification (m) = `("lenght of image"(l.))/("lenght of object"(l))`
Give: length of object, l = `(f)/(3)`
Let, l be the length of image, then, `m=(l.)/(l)=(l.)/(f//3)(or)l=(mf)/(3)`
Image of one end coincides with the object. Thus, the coinciding end must be at center of curvature.
Hence, u = R = 2f
`u = u + (f)/(3)`
`u=u.(f)/(3)=2f-(f)/(3)=(5f)/(3)`
`v=u+(f)/(3)+(mf)/(3)=(5f)/(3)+(f)/(3)+(mf)/(3)=(f(6+m))/(3)`
Mirror equation, `(1)/(v) + (1)/(u) = (1)/(f)`
`(1)/(-((f(6+m))/(3)))+(1)/(-((5f)/(3)))=(1)/(-f)`
After equation, `(3)/(f(6+m))+(3)/(5f)=(1)/(f),(3)/((6+m))=(2)/(5)`
`6 + m = (15)/(2) , m = (15)/(2) - 6`
`m = (3)/(2) = 1.5`