PLASES HELP ASAP

Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i

koban [17]

Answer:

A

$N = 1340.86 \ slits / cm$

B

$\theta = 15.7^o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 650 \ nm = 650 *10^{-9} \ m$

The angle of first bright fringe is $\theta = 5^o$

The order of the fringe considered is n =1

Generally the condition for constructive interference is

$dsin (\theta ) = n * \lambda$

=> $d = \frac{1 * 650 *10^{-9 }}{ sin(5)}$

=> $d = 7.458 *10^{-6} \ m$

Converting to cm

$d = 7.458 *10^{-6} \ m = 7.458 *10^{-6} * 100 = 0.0007458 \ cm$

Generally the number of grating pre centimeter is mathematically represented as

$N = \frac{1}{d}$

=> $N = \frac{1}{0.0007458}$

=> $N = 1340.86 \ slits / cm$

Considering question B

From the question we are told that

The first wavelength is $\lambda_1 = 650 \ nm = 650 *10^{-9} \ m$

The second wavelength is $\lambda_2 = 429 \ m = 420 *10^{-9 } \ m$

The order of the fringe is $n = 2$

The grating is $N = 5000 \ slits / cm$

Generally the slit width is mathematically represented as

$d = \frac{1}{N }$

=> $d = \frac{1}{ 5000 }$

=> $d = 0.0002 \ c m = 2.0 *10^{-6} \ m$

Generally the condition for constructive interference for the first ray is mathematically represented as

$d sin(\theta_1) = n * \lambda_1$

=> $\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}]$

=> $\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_1 = 40.5 ^o$

Generally the condition for constructive interference for the second ray is mathematically represented as

$d sin(\theta_2) = n * \lambda_2$

=> $\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}]$

=> $\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_2 = 24.8 ^o$

Generally the angular separation is mathematically represented as

$\theta = \theta_1 - \theta_1$

=> $\theta = 42.5^o - 24.8^o$

=> $\theta = 15.7^o$

4 0

2 years ago

A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially relea

grandymaker [24]

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

$A = 1.803 ft$

Period = $\frac{\pi}{2}$ seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = $\frac{w}{g}$

m= $\frac{32}{32}$

= 1 slug

Also from Hooke's Law

2 k = 32

k = $\frac{32}{2}$

k = 16 lb/ft

Using the function:

$\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0$

$x(0) = -1$ (because of the initial position being above the equilibrium position)

$x(0) = -6$ ( as a result of upward velocity)

NOW, we have:

$x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)$

However;

$x(0) = -1$ means

$-1 =c_1\\c_1 = -1$

$x(0) =-6$ also implies that:

$-6 =4(c_2)\\c_2 = - \frac{6}{4}$

$c_2 = -\frac{3}{2}$

Hence, $x(t) =-cos4t-\frac{3}{2} sin 4t$

$A = \sqrt{C_1^2+C_2^2}$

$A = \sqrt{(-1)^2+(\frac{3}{2})^2 }$

$A=\sqrt{\frac{13}{4} }$

$A= \frac{1}{2}\sqrt{13}$

$A = 1.803 ft$

Period can be calculated as follows:

= $\frac{2 \pi}{4}$

= $\frac{\pi}{2}$ seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

$x* \frac{\pi}{2} = 4 \pi$

$x \pi = 8 \pi$

$x=8$ cycles

5 0

2 years ago

It is important to wear protective equipment in both practice and games.

Ede4ka [16]

Answer:

OT

It is important to wear protective equipments during games and practices.

Explanation:

Practice is a review of what is about to happen in some time to come and with practice ,the future can be trusted.

During practice it is very important to use protective gears or equipments because without them, injuries and accidents are liable to occur.

During the major game it is also important to prevent accidents or big injuries as the equipments protect each vital part of the body.

4 0

2 years ago

In a single-slit diffraction experiment, a coherent light source illuminates a slit in a barrier, and the resulting pattern is p

UkoKoshka [18]

Answer:

hsjdsdddwqdqwdd

Explanation:

4 0

2 years ago

A bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes.

zaharov [31]

Given that,

Frequency emitted by the bat, f = 47.6 kHz

The speed off sound in air, v = 413 m/s

We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m$

or

$\lambda=8.67\ mm$

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.

3 0

2 years ago