Answer:
Because the light reflects multiple times until it gets to the Cassegrain focus.
Explanation:
The Cassegrain design can be seen in a reflecting telescope. In this type of design the light is collected by a concave mirror, and then intercepted by a secondary convex mirror, and sends it down to a central opening in the primary mirror (concave mirror), in which a detector is placed (Cassegrain focus)
Since, the light is reflected many times due to Cassegrain design, that leads to shorter telescopes.
A beam of laser is directed at a reflecting surface put on the moon when the beam of laser is reflected a receiver on the each measure the time since the beam was sent till it was received. Laser is simply light so it has constant velocity in vacuum ~ air (c = 3 x 10^8 m/s)
to find the distance:
t : time measured between launching the beam and receiving it
d : distance
d = ct
Answer:
Hi Carter,
The complete answer along with the explanation is shown below.
I hope it will clear your query
Pls rate me brainliest bro
Explanation:
The magnitude of the magnetic field on the axis of a circular loop, a distance z from the loop center, is given by Eq.:
B
= NμοiR² / 2(R²+Z²)³÷²
where
R is the radius of the loop
N is the number of turns
i is the current.
Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense, and the fields they produce are in the same direction in the region between them. We place the origin at the center of the left-hand loop and let x be the coordinate of a point on the axis between the loops. To calculate the field of the left-hand loop, we set z = x in the equation above. The chosen point on the axis is a distance s – x from the center of the right-hand loop. To calculate the field it produces, we put z = s – x in the equation above. The total field at the point is therefore
B
= NμοiR²/2 [1/ 2(R²+x²)³÷² + 1/ 2(R²+x²-2sx+s²)³÷²]
Its derivative with respect to x is
dB
/dx= - NμοiR²/2 [3x/ (R²+x²)⁵÷² + 3(x-s)/(R²+x²-2sx+s²)⁵÷²
]
When this is evaluated for x = s/2 (the midpoint between the loops) the result is
dB
/dx= - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷² - 3(s/2)/(R²+s²/4)⁵÷²
] =0
independent of the value of s.
