Ask question

Ask question

All categories

3 years ago

15

Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and th

e distance between constructive fringes on a screen 5 m away is 0.3 cm. What is the wavelength?

1 answer:

Alik [6]3 years ago

8 0

Answer:

The wavelength of the light is $7200\ \AA$ .

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes $\beta= 0.3\ cm$

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

$\beta=\dfrac{D\lambda}{d}$

Put the value into the formula

$0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}$

$\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}$

$\lambda=7.2\times10^{-7}\ m$

$\lambda=7200\ \AA$

Hence, The wavelength of the light is $7200\ \AA$ .

You might be interested in

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

7 0

3 years ago

12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

$v_{f} = v_{o}+a\cdot (t-t_{o})$ (1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v_{f}$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t_{o}$ - Initial time, measured in seconds.

$t$ - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust ( $v_{o} = 0\,\frac{m}{s}$ , $a = 30\,\frac{m}{s^{2}}$ , $t_{o} = 0\,s$ , $0\,s\le t< 30\,s$ )

$v = 30\cdot t$ (2)

Free fall ( $v_{o} = 900\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s}$ , $t_{o} = 30\,s$ , $30\,s \le t \le 120\,s$ )

$v = 900-9.81\cdot (t-30)$ (3)

Now we created the graph speed-time, which can be seen below.

5 0

3 years ago

PLSS HELP ITS TIMED <br> WILL MARK BRAINLIEST

dlinn [17]

Answer:

tjfd

Explanation:

tuff

4 0

3 years ago

g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide

vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring $\left ( m_1\right )=0.5 kg$

Compression in the spring $\left ( x\right )=20 cm$

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block $\left ( m_1\right )$

$\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2$

$v=x\sqrt{\frac{k}{m_1}}$

now conserving momentum

$m_1v=\left ( m_1+m_2\right )v_0$

$v_0=\frac{m_1}{m_1+m_2}v$

where $v_0$ is the final velocity

3 0

3 years ago

how far can a mother push a 20.0kg baby carriage, using a force of 62.0N at angle of 30.0°to the horizontal, if she can do 2920

enot [183]

Here is my solution using my app.

4 0

3 years ago