All categories

2 years ago

450 nm of light falls on a single slit of width 0.30 mm. What is the angular width of the central diffraction peak

Physics

1 answer:

Andre45 [30]2 years ago

4 0

Angular width is 3 x 10^-3

Let D be the distance between source and screed d the distance between coherent source then for central diffraction maxima,

where λ is wavelength

Given:

λ = 450 nm = 450×10^−9m

d = 0.3x10^−3m, D = 1m

W = 2 x 450×10^−9/0.3x10^−3*1

To Find:

Angular width

Solution: The width of the central maxima is nothing but the difference between the positions of the first two minima. Hence we will use the expression for the position of minima and accordingly obtain the expression of the width of central maxima and secondary maxima

θ = W/D

θ = 2 x 450×10^−9/0.3x10^−3*1/1 = 3 x 10^-3

Hence, angular width is 3 x 10^-3

Learn more about Angular width here:

brainly.com/question/25292087

#SPJ4

You might be interested in

Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows

Vlada [557]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle $\theta _c = sin^{-1} (\frac{n}{n_{slab}} )$

$sin \theta _ c =\frac{n}{n_{slab}}$

According to Snell's law of refraction:

$n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)$

At point P ; $90 - \theta _2 \leq \theta _c$

$\theta _2 = 90 - \theta _c$

Therefore:

$n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}$

Then maximum value of refractive index n of the fluid is:

$n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }$

$n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }$

n = 1.4266

3 0

3 years ago

The dances created and performed collectively by the ordinary people.

Readme [11.4K]

Answer:

Folk dance

Explanation:

hope i helped :)

6 0

3 years ago

A proton is trapped in a one-dimensional well that is 200 pm wide. What is the ground state energy of the proton, if the potenti

Harrizon [31]

Answer:

E=1.50\times 10^{-18}J

Explanation:

Energy of the one dimensional infinite well,

$E=\frac{n^{2}h^{2} }{8mL^{2} }$

Given that, $L=200 pm\\L=200\times 10^{-12}m$

For the ground state n=1,

Therefore energy is,

$E=\frac{(6.626\times 10^{-34})^{2}}{8(9.1\times 10^{-31})(200\times 10^{-12}) ^{2} }\\E=1.50\times 10^{-18}J$

8 0

3 years ago

Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

$v^2-u^2=2as$