A lens of focal length 5 cm is being used by a student in the laboratory as a magnifying glass. His least distance of distinct vision is 25 cm. What magnification is the student getting?

howw?? solve it completely......plzzz

See i m also confused how it comes -ve when it is given that it is a magnifying glass......

but see what happens: m solving this :-)

It is a magnifying glass hence the lens used is convex

Now, f = 5cm, u = -25cm, v = ?, m = ?

Using the formula i/v - 1/u = 1/f, we get v = 25/4

Now, as u know m = v/u, so putting the value of v & u in this, we get m = -0.25

 Here,

f= 5cm(Since its a convex lens)

v= 25cm (Image distace for eyes)

u=? (Object distance)

m=? (magnification)

Using lens formula,

1/v - 1/u=1/f

=> 1/v-1/f=1/u

=>1/25-1/5=1/u

=> 1-5/25=1/u

=> -4/25=1/u

=> -25/4=u

Now, Magnification m=v/u

=>m=25/-25/4=25*(-4/25)= -4

Required Answer---> The object's image is four times enlarged than object itself

no actually u r write . the magnifiacation is 0.25........in lens case magnification is = v/u....

v= 6.25 and u = 25......When we divide it we get....... 6.25/25 = +0.25.........Thnxxxx a lot.....for helping me

What do u think ??????? Is it correct ??????  I think I have solved it according to the Q.........

  Here,

f= 5cm(Since its a convex lens)

v= -25cm (Image distace for eyes)

u=? (Object distance)

m=? (magnification)

Using lens formula,

1/v - 1/u=1/f

=> 1/v-1/f=1/u

=> -1/25-1/5=1/u

=> -1-5/25=1/u

=> -6/25=1/u

=> -25/6=u

Now, Magnification m=v/u

=>m= 25/-25/6

=> m=25*6/-25

=> m= -6

Required Answer---> The object's image is six times enlarged than object itself

Hey King of merit.......... how there will be v = 25 cm ???????? Ist thing it is object distance ofcourse because image distance 4 humans is 2.5cm (4rm lens to retina) & the 2nd thing is that it'll be in -ve............. plz explain !!!!!!

@Shubham: I know m=v/u in mirror formula but see here's u is given in negation....... 

 @ king of merit...... u is already given.......we have to find v............

Hey Subham did u get any idea about this ??????

 Its the same type of question as we got in the calculation of lens power.

here we will calculate object distance taking the childs normal vision as image distance. We will find that the child's near point shift closer to his eyes and hence the object distance would we more closer than 25cm. Since it forms a real image its magnification would always be negative. ANd the same way has been adopted by me and i got -25/6cm as his near point and -6 as his magnification

 @ apoorwa - no.,,, 

 Follow this link---> http://cbse.meritnation.com/discuss/question/794241

and yes the eyes vision is always taken as v and not u

@ kinf of merit - in this link http://cbse.meritnation.com/discuss/question/794241 it is given that object distance is 24 but in case of convex lens used in correction of hypermetropia u is always 25 cm....

No......... I m totally sure that if we are taking about humans then the image distance is 2.5cm which is the distance between eye lens and retina ........ It always remains constant so, v = 2.5 cm 4 all conditions :-)

 anyone der or lost in the link???

v or u?? 

So, what do u say is my answer wrong or right??

 Hey guys i'm going offline.....If u want any further discussion give me your Id.........

 First of all everyone think of one thing! Its a convex lens and if it's kept within a rangeable distance would it not produce a m>1 but see apoorwa's answer its 0.25, tht too in +ve sign which is impossible for convex lens

@ king of merit: See i m not confident but acc. 2 me u r wrong :-(

@ shubham: v is 2.5cm and m absolutely sure about that

bye gtg now but i'll be back in about 1 hour.............. i'll try 2 find ur answer as m too in suspence

but if u'll get that then plz inform me :-)

@ King of merit: yes, that's what m thinking about but this is the only method acc. 2 me as we can't take v=25cm !!!!!!!!

Ok apoorwa I’ll inform u about the answer but I think u r right ………ok byee..

Thnxxxxx  a lot

@ Subham: I think I got that actually we all were wrong becoz this was the condition of two lenses :eye lens & magnifying glass. So, we have to use formula 4 both of them :-)

@ shubham  ,  apoorwa ,  guru....c  guys  u  oll  wr  pondering  ovr  a  silly  question....its  guru's  answer  which  is  the  correct  one  !

1. 1st  of  oll  the  eye's  distance  are  olways  taken  as  v  ..!  n  in apoorwa's  case  v  ws 6.25  which  is  impossible  for  our  eye  to  view !

 2.  the  hight  of  the  image  must  be  larger  than  the  object's  thus  its  guru's  answer  the  sensible  one

 hey guys i have the answer..

in this case magnification by a lens and angular magnification by an optical instrument are two different things.

Angular magnification = 25 /6.25

Therefore , angular magnification = 25/u ( 25 cm is the near point)

  = 25/-6.25

  =  -4

So the magnification will be 4……