11thank you bhaiya...
1) When an object is at a distance x1 and x2 from the poles of a concave mirror , imagesm of same magnification are formed. The magnitude of the focal length of the mirror is :
a) \left|\frac{x_{1}+x_{2}}{2} \right| b)\left|\frac{x_{1}-x_{2}}{2} \right| c)\left|x_{1}-x_{2} \right| d))\left|x_{1}+x_{2} \right|
ans a
whatever method i'm tryin i'm ending up with x1 = x2. Pls help....
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2 Answers
106there are two cases when the magnification (in magnitude) can be same . One is when it is between f and c and other is when object is between pole and f.
1/v + 1/(-x1) = 1/(-f) .. (1)
=> 1/v = (f - x1)/(x1.f)
=> v = (x1.f)/(f - x1)
|m| = |v/u| = |f|/|f - x1| = |f|/|f - x2|
=> |f - x1| = |f - x2|
now we know that one of x1,x2 is less than f and other is greater than f
=> f - x1 = -(f - x2)
=> x1 + x2 = 2f
=> f = |(x1 + x2)/2|