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 12cm 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?
Answers
Answered by
136
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 cm
f=8cm
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.
1/v-1/u=1/f
u= -12 cm
v= 24 cm
1/f=(1/24)-1(-/12)
1/24+1/12=
1/8 cm
f=8cm
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.
vinodkumar:
hope this helps you
pls mark as the best
Similar questions