Someone please help me!

A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick

DerKrebs [107]

Dispersion angle = 0.3875 degrees.
Width at bottom of block = 0.09297 cm
Thickness of rainbow = 0.07038 cm
Snell's law provides the formula that describes the refraction of light. It is:
n1*sin(Î¸1) = n2*sin(Î¸2)
where
n1, n2 = indexes of refraction for the different mediums
Î¸1, Î¸2 = angle of incident rays as measured from the normal to the surface.
Solving for Î¸2, we get
n1*sin(Î¸1) = n2*sin(Î¸2)
n1*sin(Î¸1)/n2 = sin(Î¸2)
asin(n1*sin(Î¸1)/n2) = Î¸2
The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
Red:
asin(n1*sin(Î¸1)/n2) = Î¸2
asin(1.00029*sin(40.80)/1.641) = Î¸2
asin(1.00029*0.653420604/1.641) = Î¸2
asin(0.398299876) = Î¸2
23.47193844 = Î¸2
Violet:
asin(n1*sin(Î¸1)/n2) = Î¸2
asin(1.00029*sin(40.80)/1.667) = Î¸2
asin(1.00029*0.653420604/1.667) = Î¸2
asin(0.39208764) = Î¸2
23.08446098 = Î¸2
So the dispersion angle is:
23.47193844 - 23.08446098 = 0.38747746 degrees.
Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
tan(Î¸) = X/11.6
11.6*tan(Î¸) = X
So for Red:
11.6*tan(Î¸) = X
11.6*tan(23.47193844) = X
11.6*0.434230136 = X
5.037069579 = X
And violet:
11.6*tan(Î¸) = X
11.6*tan(23.08446098) = X
11.6*0.426215635 = X
4.944101361 = X
So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm
The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
cos(Î¸) = X/0.092968218
0.092968218*cos(Î¸) = X
0.092968218*cos(40.80) = X
0.092968218*0.756995056 = X
0.070376481 = X
So the distance between the red and violet rays is 0.07038 cm.

7 0

3 years ago

In this experiment you will investigate which of the following properties of Faraday's law of electromagnetic induction? (Select

11Alexandr11 [23.1K]

Answer:

Answered

Explanation:

Part A

According to Faraday's law the induced emf in coil is equal to negative of its rate of change of magnetic flux time the number of turns in the coil.

$\epsilon = -N\frac{d\phi}{dt}$ = $-N\Delta\frac{BA}{\Delta t}$

When an emf generated by a change of magnetic flux, produced current of whose magnetic field opposes the change which produces it.

By the above equation the correct options are 1,2 and 4

Part B

Large signals of frequency of 60Hz are measured by osciloscope.

Hence the correct option is part 1.

3 0

3 years ago

Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0

3 years ago

The circuit shown in the figure (Figure 1) uses a neon-filled tube. This neon lamp has a threshold voltage V0 for conduction, be

Svetllana [295]

Part 1)

Answer:

Explanation:

As we know by equation of charging of the capacitor we will have

$V = E(1 - e^{-t/RC})$

so we will have

$87 = 105(1 - e^{-t/RC})$

here we know that

$R = 3.00 \times 10^6 ohm$

$C = 0.250 \mu F$

so we have

$t = 1.32 s$

Part b)

Answer:

The time will increase.

Explanation:

As we know that on increasing the value of the resistance the the product of the resistance and capacitance will increase so the time will increase to get the above voltage.

Part c)

Answer:

The capacitor discharges through a very low resistance (the lamp filled with ionized gas), and so the discharge time constant is very short. Thus the flash is very brief.

Explanation:

Since the lamp resistance is very small so the energy across the lamp will totally lost in very short interval of time

Part d)

Answer:

Once the lamp has flashed, the stored energy in the capacitor is gone, and there is no source of charge to maintain the lamp current. The lamp "goes out", the lamp resistance increases, and the capacitor starts to recharge. It charges again and the process will repeat.

Explanation:

Since we know that the battery is connected to the given system so after whole energy of capacitor is flashed out it is again charged by the battery and the process will continue

3 0

4 years ago

The distance between Pluto and the Sun is 39.1 times more than the distance between the Sun and Earth. Calculate the time taken

german

Answer:

Explanation:

Given

Distance between Pluto and sun is 39.1 times more than the distance between earth and sun

According to Kepler's Law

$T^2=kR^3$