With light microscopy, if the objective lens (lens closest to the specimen) magnifies 40-fold, and the eyepiece lens magnifies 1
1 answer:
The final magnification will be 400-fold or 400 times the original size of the object.
For magnifying smaller objects, a compound microscope is used.
A compound microscope consists of an objective and an eyepiece, whose diagram is shown in the adjoining image.
The lens near the object is called an objective and the other one is the eyepiece.
Let the magnification of the objective be m1
Let the magnification of the eyepiece be m2
The final magnification by the microscope, M, will be
M = m1 x m2
Putting the values in the above equation
M = 40 x 10
M= 400
Thus, the final magnification will be 400-fold or 400 times the original size of the object.
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Answer:
Options (3) and (6)
Explanation:
Janine drives 2.2 km East.
Since, distances measured in the East are positive,
Displacement = 2.2 km
Then she drives 4.4 km west
Displacement = -4.4 km
Followed by 1.7 km in the East
Displacement = 1.7 km
Total displacement = 2.2 - 4.4 + 1.7 = -0.5 km
km
Total distance covered by Janine = 2.2 + 4.4 + 1.7 = 8.3 km
d = 8.3 km
Therefore, Options (3) and (6) will be the answer.
Answer:
the Zeeman effect without spin is three lines
Explanation:
The Zeeman effect is the result of the interaction of the magnetic field with the orbital angular momentum of the electrons, if we do not take the spin into account it is called the Normal Zeeman effect.
Therefore we only take into account the orbital moments (m_l) of the transition, from the selection rules of the refreshed harmonics, only the transition with
= 0, ± 1
ΔE
Let's analyze for the case of the Hydrogen atom
For a transition between levels n = 1 and n = 2 the values of m_l are n_f = 1 m_l = 0 and for n₀ = 2 m_l = 0, 1
so we only have two lines.
For transition n_f = 2 and n₀ = 3
n_f = 2 m_l = 0, 1
n₀ = 3 m_l = -1, 0, 1
There are only two lines plus the central line, so there are three spectral lines
for n_f = 3 and n_o = 4
n_f = 3 ml = -1, 0, 1
n₀= 4 ml = -2, -1, 0, 1, 2
Only transitions with Δm_l = ±1 are allowed, so there are only two transitions plus the central transition (Δm_l = 0), so there are only 3 spectral lines.
In summary, due to the selection rule of spherical harmonics, the greatest number of lines in the Zeeman effect without spin is three lines.