✔ 最佳答案
08.32. The answer is D. 150 cm
For the telescope to be in normal adjustment, the focus of objective lens and eyepiece must coincident with each other, with the focal length of objective being much larger than that of eyepiece.
So, eyepiece has a focal length of 2 cm = 0.02 m
Objective lens has a focal length of 1 m
By 0.02 / 1 = D / 0.03
We have diameter of objective lens, D = 1.5 m = 150 cm
04.12. The answer is B. -3V
For the charge +Q at A, potential, V = kQ / (2r)
where r is the distance AC or CB, k is a constant
Now, for the charge -2Q at C, potential at B, V' = k(-2Q) / r = -4V
And the final electric potential at B is the algebraic sum of the two amounts, V + V' = V - 4V = -3V
00.20. The answer is D. (2) and (3) only.
The ray OS is undeflected when it passes through L since it passes through the optical centre.
And PQ, for the left side, it must passes through the focus of L.
So, the rays are converged. L is a convergent lens (convex lens)
(1) is incorrect.
For (2), if you draw the ray-diagram, you will find the point must lie along the line OS, since OS is undeflected.
For (3), the image cannot be formed on the right side of L, so the image must be virtual.
84.13. The answer is B. II.
For P and Q, when they meet together at III, the image formed there, it is on the focal plane.
So, we can determine that II is the focus of the lens.
For converging lens, when a ray XY which is parallel to the plane of the lens passes through L, the refracted ray must passes through the focus, which is II.
2009-07-13 09:27:49 補充：
08.32. This is already the calculation...
2009-07-13 09:30:32 補充：
04.12. Sorry for the mistake. That is 04 AS physics
12. B
Blue light refracted more than red light in the lens. So, P is the red light
As the light rays are converging, so the lens is a concave lens
Q should travel slower than P in the lens as Q is refracted more. You can prove it by Snell's law.
2009-07-14 09:21:06 補充：
You can find it by simple geometry, by the assumption that they have the same focal power.
