The image of a small blue ball placed 50 cm from a convex lens with a focal length of 10 cm is calculated using the lens equation. By applying the formula 1/f = 1/di + 1/do, where do is the object distance and di is the image distance, the values are substituted to find di. After calculations, it is determined that the image distance (di) is 25 cm. Therefore, the image of the blue ball will be formed 25 cm behind the lens. This demonstrates the application of the thin lens formula in optics.
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athenaroa
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A very small blue ball is placed 50cm to the left of a convex lens of height 5 cm and focal length 10 cm. how far (in cm ) behind the lens will the image of the ball be formed.
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler: