Question

a thin rod of length f/3 is placed along the optic axis of a concave mirror of focal length f such that the image which is real and elongated just touches the rod calculate the magnification produced by the mirror

Answer 1

Since the image touches the rod, one end of the rod must be at C of the mirror. 
Since the image is real and elongated, the other end of the rod must be between C and F of the mirror. 
The distance of this latter ( A) end from the optic center of the mirror is (R-f/3) = 2f-f/3 
= 5f/ 3 
(u –f)= 5f/ 3-f = 2f/3 
m = f/ (u-f) = 3/2 = 1.5 

Magnification of the A end of the rod is 1.5 where as the magnification of the joint of image and rod is 1. 



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Answer 2

Answer:

1.5

Explanation:

length of rod = f/3

as the image is formed just touching the rod. it means that that point will be the centre of curvature.

so distance from pole will be 2f

Now,

distance of object from mirror = 2f - f/3 = 5f/3

Therefore, u = - 5f/3 ( negative sign as object is to left)

By mirror formula,

$\frac{1}{v } + \frac{1}{u } = \frac{1}{f}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{v} = \frac{u - f}{fu}$

$v = \frac{fu}{u - f}$

focal length of concave mirror is -f

$v = \frac{ - f \times - 5 \frac{f}{3} }{ - 5 \frac{f}{3} - ( - f)} = \frac{ \frac{5 {f}^{2} }{3} }{ \frac{ - 2f}{3} } = \frac{ - 5f}{2}$

$magnifiction = \frac{v}{u} = \frac{ \frac{ - 5f}{2} }{ \frac{ - 5f}{3} } = \frac{3}{2} = 1.5$