A) \[(n+1)f\]
B) \[(n-1)f\]
C) \[\left( \frac{n+1}{n} \right)f\]
D) \[\left( \frac{n-1}{n} \right)f\]
Correct Answer: D
Solution :
As the magnification, \[m=\frac{\upsilon }{u}\] .....(i) As the lens formula is \[\frac{1}{\upsilon }-\frac{1}{u}=\frac{1}{f}\] \[\frac{u-\upsilon }{u\upsilon }=\frac{1}{1}\] \[\Rightarrow \] \[uf\left( 1-\frac{\upsilon }{u} \right)=uV\] \[\Rightarrow \] \[1-m=\frac{u\upsilon }{uf}\] \[\Rightarrow \] \[m=1-\frac{u\upsilon }{uf}\] \[m=\frac{f-\upsilon }{f}\] ??..(ii) Now from equations (i) and (ii), we get, \[\Rightarrow \] \[\frac{1}{n}=\frac{-f+\upsilon }{-f}\] \[\upsilon =\left( \frac{n-1}{n} \right)\,f\]
You need to login to perform this action.
You will be redirected in
3 sec
