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?

** 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**

- -2

** 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**

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**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....... **

- 10

** 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**

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** Foll**o

*w this link--->*

*http://cbse.meritnation.com/discuss/question/794241**and yes the eyes vision is always taken as v and not u*

- -1

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

**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**

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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……

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