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Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

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

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

or

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0

2 years ago

What are the wavelength ranges for the following? (a) the AM radio band (540–1600 kHz) maximum wavelength m minimum wavelength m

Pie

Answer:

Explanation:

a ) AM radio band (540–1600 kHz)

frequency = 540 kHz = 540 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 540 x 10³

= 555.55 m

frequency = 1600 kHz = 1600 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 1600 x 10³

= 187.5 m

maximum wavelength = 555.55 m

minimum wavelength = 187.5 m

b )

AM radio band (88 - 108 MHz)

frequency = 88 MHz = 88 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 88 x 10⁶

= 3.41 m

frequency = 108 MHz = 108 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 108 x 10⁶

= 2.78 m

maximum wavelength = 3.41 m

minimum wavelength = 2.78 m

3 0

2 years ago

What is the instantaneous velocity v of the particle at t=10.0s?

algol [13]

The instantenous velocity is just the slope of the graph at a certain instant. Since the graph is a straight line, its instantenous velocity is uniform through out. v = dx / dt = (40 - 10) / (50 - 0) = 0.6 m/s.

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6 0

3 years ago

Read 2 more answers

In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y

brilliants [131]

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

$E=mc^{2}$

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

$Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules$

Thus the mass that is covertred at 100% efficiency is

$mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg$

Part 2)

At 30% efficiency the mass converted equals

$mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg$

8 0

3 years ago

Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2.

IrinaVladis [17]

Answer:

option C

Explanation:

Let mass of the bullet be m and velocity be v

mass of gun be M and bullet be V

now,

using conservation of momentum for gun 1

(M+m) V' = 2 mv + 3 MV

V' = 0

3 M V = - 2 mv

momentum of gun 1 =- 2 mv---------(1)

now for gun 2

(M+m) V' = mv + MV

V' = 0

M V = - mv

momentum of gun 1 = -mv-----------(2)

dividing equation (1) by (2)

$\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}$

$\dfrac{P_m1}{P_m2} = \dfrac{2}{1}$

the correct answer is option C

7 0

2 years ago