Question

I NEED THE ANSWER NOW..PLS.. A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens,in which way will he have to move the screen to get a sharp image of the object on it again? How will the magnification of the image be affected?

Answer 1

Lens formula: 1/v - 1/u = 1/f

u = -12 cm
v = 24 cm

1/f = (1/24)     -    1/(-12)
      = 1/24  + 1/12
      = 1/8

f = 8 cm.

When u = -12 cm, the object is placed between F and 2F and the image is formed beyond 2F on the other side. This is a enlarged inverted image. If the object is moved away from the lens, the image moves towards the lens. So the screen should be moved towards the lens.

Magnification m = v/u = 24/(-12) =  -2

If the denominator increases, v decreases and hence the magnification decreases. So the image is smaller than the previous image. 
   

Answer 2

Given:- A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens,in which way will he have to move the screen to get a sharp image of the object on it again? How will the magnification of the image be affected?⤵

Solution:-⤵

From the lens formula,

=

v

1

−

u

1

=

f

1

⟹

f

1

=

24

1

−

−12

1

=

8

1

⟹f=8

When u=−12, the object is placed between F and 2F and the image is formed beyond 2F on the other side. This is a enlarged inverted image. If the object is moved away from the lens, the image moves towards the lens. So the screen should be moved towards the lens.

Magnification=

u

v

=

−12

24

=−2

When denominator increases, v decreases and hence magnification decreases.