Question

A student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that 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 screen ?
B) What happens to the size of the image of the flame formed on the screen ?
C) What difference is seen in the intensity (brightness) of the image on the screen ?
D) What is seen on the screen when the flame is very close (at about 5 cm) to the lens ?

Answer 1

A)in the negative x axis he should move the lens.B)since the object distance decreases so the image formed will be illuminated and its height will vary depending upon the position of the object. C)the intensity of the image will decrease as distance WL increase.becuz distance is inversely proportional to the intensity of the light.
D)the image formed will be at a distance of 10 cm from the lens.also the magnification=2.the image wl be illuminating

Answer 2

A) 1/v = 1/f + /1u ,  u is negative.  let it be  -U.
           = 1/f - 1/U
    U is decreased. 1/U increases.  So RHS decreases.  So 1/v decreases.
    So v increases.
    Thus the lens is moved away from screen, towards the object.

B)  As v increases and u decreases,  magnification increases.
      So image is magnified.

C)  As the image is enlarged, the light rays are spread out.
      So the brightness or intensity of the image is reduced.

D) When the flame is at Focal length distance, the image is at Infinity.
     If the flame is closer than focal length, then a virtual erect image
     is formed.
  
       1/v = 1/f +1/u = 1/10 - 1/5 = -1/10
        v = 10 cm,  a virtual image is formed.
       magnification = v/u = -10/-5 = 2    So image is erect.