Question

a student focused the image of a candle flame placed at about 2m from a convex lens of focal length 10 cm , on a screen. After he moves gradually the flame towards the lens and each time focuses its image on the screen.

A) In which direction does he move the lens to focus the flame on the screen

Answer 1

for convex lens ,

focal length = 10 cm
distance between candle and lens = -2m

use formula,
1/v - 1/u = 1/f

1/v - 1/(-200) = 1/(10)
1/v = 1/10 - 1/200
=19/200

v = 200/19 cm

hence, screen placed right side of lens ,

let distance distance between object and screen = D

and lens slide between in such a way that , image always form on screen .
let at x distance from object put the lens then it's form image on screen .

so, distance of object form lens = -x
distance of image from lens= D-x
focal length = -f

use formula,

1/v - 1/u = 1/f

1/(D-x) - 1/(-x) = 1/f

1/(D -x) + 1/x = 1/f

{ x + D - x }/(D -x)x = 1/f

Df = Dx -x²

x² +Dx -Df

x = { -D ± √(D²-4Df )}/2

only two positions exist where we will put the lens then lens form image in screen .

these are ,

x = ( -D -√(D²-4Df) )/2 , and x = ( -D+√(D²-4Df) )/2

case 1:- the lens can be moved towards the object to form an image on the screen ,
image is real and inverted and enlarged.
the image is close to the focus .
this case u > v ,
e.g u decrease and v increase.

case 2:- when lens is move towards the screen .
in this case image distance is less then initial.
hence, u increases and v decreases .

Answer 2

For the given case,  it seems that the lens is moved towards the object or away from the screen to form real inverted image on the fixed screen.

For a biconvex lens.

Once the image is focused on the screen  with  u = -200 cm, the object is moved towards the lens (of course towards the screen).  That means object distance is reduced. Then image distance also reduces.  Hence, the lens is moved towards the screen.

We want to know the relationship between v and (u+v). 

Initially,   1/v - 1/u = 1/f
               1/v = 1/10  - 1/200 = 19/200
                v = 200/19 cm = 10.53 cm

Let the distance between the screen and the object = d = -u + v.
      (u is negative).      Initially  d = 200+10.53 = 210.53 cm

Then  u = v - d

Applying lens equation:   1/v = 1/f + 1/u = 1/f + 1/(v-d)
               1/v = (v-d+f) / ([f (v-d)
               v(v-d+f) = f(v-d)
               v² - d v + df = 0
               v = [d + √(d²-4df)] /2
There are two values of image distance for each value of d.
            v1 = [ d - √(d² - 4df) ] /2,        v2 = [ d + √(d² - 4df) ] /2

====>  calculate v for reducing values of d.  If you differentiation in calculus, you could find differential of v1 wrt d, then we find it is negative for the given values. So as d decreases, v1 increases.

When the object is moved towards the lens:
case 1) The lens can be moved towards the object to form an image
             on the screen. In this case the inverted real image is enlarged.
             The image is close to the focal point F.
                  u > v.     u decreases, v increases
 
Case 2)   In the case of v2, the lens is moved towards the screen.
                In this case the image distance is less than initial.  Hence, 
                      u increases, v decreases.

see the table for u and v.