Question : An object approaches a convergent lens from left of the lens with uniform speed 5 m/s and stops at the focus. The image
a) moves away from the lens with an uniform speed 5 m/s.
b) moves away from the lens with an uniform acceleration.
c) moves away from the lens with a non - uniform acceleration.
d) moves towards the lens with a non - uniform acceleration.
Doubt by Falguni
Solution : c) moves away from the lens with a non - uniform acceleration.
Explanation :
We already know that when object is move towards the lens the image starts moving away from the lens. For the rate of movement of the image let us consider that the focal length of the convergent lens would be 10 cm and the object is being moved towards the lens with uniform velocity of 5 cm/s then the corresponding distance of the image would be given by using the thin lens formula
1/v - 1/u = 1/f
1/v = 1/f +1/u
1/v = (u+f)/uf
v = uf/(u+f)
u = -ve
so
v = -uf/(-u+f)
v = -uf/(-u+f)
f = 10 cm
When u = 30 cm
v = -30×10/(-30+10)
v = -300/(-20)
v = 15 cm
When u = 25 cm
v = -25×10/(-25+10)
v = -250/(-15)
v = 16.67 cm
When u = 20 cm
v = -20×10/(-20+10)
v = -200/(-10)
v = 20 cm
When u = 15 cm
v = -15×10/(-15+10)
v = -150/(-5)
v = 30 cm
When u = 10 cm
v = -10×10/(-10+10)
v = -100/(0)
v = ∞ cm
| S.No. | Object Distance (u) | Image Distance (v) |
| 1. | 30 cm | 15 cm |
| 2. | 25 cm | 16.67 cm |
| 3. | 20 cm | 20 cm |
| 4. | 15 cm | 30 cm |
| 5. | 10 cm | ∞ |
From the above, it is clear that when the object distance is changed uniformly then the image distance is not charging uniformly. So we can say that image will move away from the lens with a non-uniform acceleration.
