Ask question

Ask question

All categories

2 years ago

11

"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a s

hift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?"

1 answer:

Tamiku [17]2 years ago

6 0

Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

You might be interested in

An applied force varies with position according to F = k1 x n − k2, where n = 3, k1 = 3.6 N/m3 , and k2 = 76 N. How much work is

aniked [119]

Answer:

The work done is 205 kJ.

Explanation:

Hi there!

Work can be calculated using the following equation:

W = F · Δx

Where:

W = work

F = applied force

Δx = displacement

In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:

W = ∫ F · dx

F = 3.6 N/m³ · x³ - 76 N

W = ∫ (3.6 x³ - 76)dx

W = 0.9 x⁴ - 76x

Evaluating from xi to xf:

W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m

W = 205 kJ

8 0

3 years ago

What happens when a light ray is parallel to the principal axis ?

Phantasy [73]

Answer:

Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens. ... These rays of light will refract when they enter the lens and refract when they leave the lens.

Hope this helps...

7 0

2 years ago

When you go camping, you burn wood. Are you contributing to air pollution?

Jet001 [13]

Yes because of the smoke you are creating in the air

3 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

Assume that the equator of the Earth is 24,000 miles long (that’s its circumference). Now, pretend that you are standing somewhe

slega [8]

Answer:

1000 mph

Explanation:

P = Perimeter = 24000 mi

r = Radius of the equator

t = Time taken to complete one rotation = 24 h

Perimeter of a circle is given by

$P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi$

Angular speed is given by

$\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}$

Velocity if given by

$v=r\omega\\\Rightarrow v=\dfrac{24000}{2\pi}\times \dfrac{2\pi}{24}\\\Rightarrow v=1000\ mph$

The person would be going at a speed of 1000 mph

3 0

3 years ago