|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Apr 2008 07:57:03 IST
|
|
|
Que. - When a object is placed at a distance of 60 cm. from a convex mirror, the magnification produced is 0.5. Where should the object be placed to get a magnification of 1/3?
|
Ankur
http://www.krazzyankuriit.co.cc/ |
|
|
|
13 Apr 2008 09:50:16 IST
|
|
|
from first condition ..ques. tells u abt v..& hence f..
now u have f &v...calculate for u from magnification formula and mirror formula..
|
Don't Dream ..Do the dream...
Rock On ....
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
from the first case,
applying magnification and mirror formula, we get f = +60cm.
the we use again magnification and mirror formula taking u = -3v,
we get u = -120 cm which is req answer.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Apr 2008 00:52:28 IST
|
|
|
u= -60 cm m= 0.5 m= -v/u v= -mu = -0.5*(-60) = 30 cm According to mirror formula, 1/f=1/v+1/u =1/30+(-1/60) = 60 cm New m=1/3 m= -v/u 1/3= -v/u u=-3v Again, 1/f=1/v+1/u 1/60= 1/v+(-1/3v) v= 40 cm Thus u= - (3*40) = -120 cm
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
