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.
Given: Mirror is concave
The image is real and elongated and it just touches the rod
To Find: Magnification
Solution: Let the length of rod = L
then, [tex]m=\frac{L}{\frac{f}{3} }[/tex]
[tex]L=\frac{mf}{3}[/tex]
It is understood that the Image of one end coincides with object
u' = 2f
[tex]u'=u+\frac{f}{3}[/tex]
[tex]u=2f-\frac{f}{3} = \frac{5f}{3}[/tex]
[tex]v=-(u+\frac{f}{3} +\frac{mf}{3})[/tex]
A mirror formula can be characterized as the recipe which gives the connection between the distance of object 'u', the distance of image 'v', and the focal length of the mirror 'f'.
By using Mirror Formula,
[tex]\frac{1}{u+\frac{f}{3}+\frac{mf}{3} }+\frac{1}{u}=\frac{1}{f}[/tex]
[tex]\frac{3}{5f+f+mf}+\frac{3}{5f} =\frac{1}{f}[/tex]
[tex]\frac{1}{m+6}=\frac{2}{15}[/tex]
[tex]m=\frac{3}{2}[/tex]
The magnification is 1.5
