Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is 
Explanation:
From the question we are told that
The focal length is 
This negative sign shows the the microscope is diverging light
The angular magnification is 
The distance between the objective and the eyepieces lenses is 
Generally the magnification is mathematically represented as
![m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]](https://tex.z-dn.net/?f=m%20%20%3D%20%20%5B%5Cfrac%7BZ%20-%20f_e%20%7D%7Bf_e%7D%5D%20%5B%5Cfrac%7B0.25%7D%7Bf_0%7D%20%5D)
Where 
is the eyepiece focal length of the microscope
Now making the subject of the formula
![f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }](https://tex.z-dn.net/?f=f_e%20%20%3D%20%5Cfrac%7BZ%7D%7B1%20-%20%5B%5Cfrac%7BM%20%20%2A%20%20f_o%20%7D%7B0.25%7D%5D%20%7D)
substituting values
![f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }](https://tex.z-dn.net/?f=f_e%20%20%3D%20%5Cfrac%7B%200.19%20%7D%7B1%20-%20%5B%5Cfrac%7B300%20%20%2A%20%20-0.0055%20%7D%7B0.25%7D%5D%20%7D)
Explanation:
Water (H2O) as a polar covalent molecule has its arrangement of oxygen and hydrogen atoms where one end (hydrogen) has a partially positive charge while the other side (oxygen) had a partially negative charge.
It is also capable of forming hydrogen bonds with polar molecules. Each water molecule can form two hydrogen bonds involving their hydrogen atoms and two further hydrogen bonds using the hydrogen atoms attached to neighboring water molecules.
<span>11.823 cm
There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall.
The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm.
Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So
82 + 1/21 * 2 = 82 + 2/21 = 82.0952381
Now we have the following dimensions with a circle replacing the ball in the original problem.
Distance from wall to effective circle = 82.0952381 cm
Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm
Effective diameter of circle = 3.995462279 cm
And because the geometry makes similar triangles, the following ratio applies.
3.995462279/41.9047619 = X/124
Now solve for X
3.995462279/41.9047619 = X/124
124*3.995462279/41.9047619 = X
495.4373226/41.9047619 = X
11.82293611 = X
The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
Answer:
<u>Resistance at 25 V</u>
Explanation:
(10)
(A)
10
20
100
200
Determining Current in a Parallel Circuit
Observed
Resistor Set
(0)
Total
Resistance
Calculated
Current
(A)
Current
(A)
(10)
20, 20, 20
20, 20, 200
<h3>Voltage needed to raise current to 3.75 A (20, 20, 200 resistor set):</h3>
Calculated
Observed:
Calculating Power of Circuit Components
Current through Each Bulb
(A)
Table B
Table C
Table D
Observed Total Current
(A)
Current Experimental
(A)
Observed Current
through Each Resistor
(A)
Power Usage per Bulb
(
I don't know where you live, but chances are you won't see much at all. Light pollution levels have increased so much due to street lamps and other lights that it's practically impossible to see stars in the night time with the naked eye.
