1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
GREYUIT [131]
3 years ago
8

A diffraction grating has 300 lines per mm. If light of wavelength 630 nm is sent through this grating, what is the highest orde

r maximum that will appear? A diffraction grating has 300 lines per mm. If light of wavelength 630 nm is sent through this grating, what is the highest order maximum that will appear? 5 6 2 8 5.3A diffraction grating has 300 lines per mm. If light of wavelength 630 nm is sent through this grating, what is the highest order maximum that will appear? A diffraction grating has 300 lines per mm. If light of wavelength 630 nm is sent through this grating, what is the highest order maximum that will appear? 5 6 2 8 5.3
Physics
1 answer:
True [87]3 years ago
8 0

Answer:

The order of maximum is   n = 5

Explanation:

From the question we are told that

  The  diffraction grating is  k  =  300 lines per mm  =  300000 lines per m

   The wavelength is  \lambda  =  630 \  nm  =  630 *10^{-9} \  m

   Generally the condition for constructive interference is mathematically represented as

      dsin \theta = n  * \lambda

Here n is the order maximum

d is the distance the grating which is mathematically represented as

    d =  \frac{1}{k}

=>   d =  \frac{1}{300000}

=>    d =  3.3*10^{-6}\  m

So

   n  = \frac{dsin \theta}{ \lambda}

at maximum  sin\theta  =  1

     n  = \frac{d}{\lambda}

=>   n  = \frac{3.3*10^{-6}}{630 *10^{-9}}

=>   n = 5

 

You might be interested in
The relative density of oxygen and carbon dioxide are 16, 12 respectively. If 25cm3 of carbon dioxide effuse out in 75 sec what
Mrac [35]

Answer:

32 cm³

Explanation:

The given gas data are;

The relative density of oxygen = 16

The relative density of carbon dioxide = 12

The time it takes 25 cm³ of carbon dioxide to effuse out = 75 seconds'

The duration of effusion of the oxygen = 96 seconds

The rate of effusion of carbon dioxide, R1 = 25 cm³/(75 sec) = (1/3) cm³/sec

According to Graham's law of diffusion and effusion of a gas, we have;

\dfrac{Rate \ of \ effudion \ of \ gas \ 1}{Rate \ of \ effudion \ of \ gas \ 2} =\dfrac{The \ relative \ density \ of \ gas \ 2}{The \ relative \ density \ of \ gas \ 1}

Therefore, we have;

\dfrac{Rate \ of \ effudion \ of \ oxygen}{(1/3)} =\dfrac{12}{16}

The \ rate \ of \ effudion \ of \ oxygen}=\dfrac{12}{16} \times \left(\dfrac{1}{3 } \ cm^3/sec\right ) = \dfrac{1}{4} \ cm^3/sec

The volume of effusion = The rate of effusion × Time

The volume of the oxygen that will effuse in 96 seconds is given as follows;

The rate of effusion of a gas × Time

V = The rate of effusion of oxygen × Time = (1/3) cm³/sec × 96 sec = 32 cm³

The volume of oxygen that will effuse in 96 seconds, V = 32 cm³.

8 0
3 years ago
What is the relationship between Potential energy and KInetic energy in a system?
steposvetlana [31]

Answer:

A soma dos dois tipos de energia, potencial e cinética, resultam na energia mecânica. Ou seja, ambas compõe a energia mecânica.

8 0
3 years ago
(eText prob. 4.25 with some values changed) Air enters a diffuser of a jet engine operating at steady state at 2.65 psia, 389◦R,
krek1111 [17]

Answer:

V_2 = 45.44m/s

Explanation:

We have to many data in different system, so we need transform everything to SI, that is

P_1 = 2.65 Psi = 18.271 kPa\\T_1= 389\°R = 216 K\\V_1 = 869ft/s = 264m/s\\T_2 = 450\°R = 250K

When we have all this values in SI apply a Energy Balance Equation,

\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0

Solving for V_2

V_2 = \sqrt{V_1^2+2(h_1-h_2)}

From the table of gas properties we calculate for T_1 = 216K and T_2 = 250K

h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})

h_1 = 215.97kJ/kg

For T_2;

h_2 = 250.05kJ/kg

Substituting in equation for V_2

V_2 = \sqrt{V_1^2+2(h_1-h_2)}

V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}\\V_2 = 45.44m/s

4 0
3 years ago
which surface would have the greatest friction? low carpet a gym floor a brand new sidewalk or gravel​
jasenka [17]
Carpet I’m guessing
7 0
3 years ago
A 10 lb. cat needs 31⁄2 tablets that each contain 6 mg of a drug. what's the equivalent dose in mg/kg?
mina [271]

Mass of cat = 10 lb

1 lb = 0.454 kg

So mass of cat is 4.54 kg

Number of tablets given to cat

N = 3\frac{1}{2}

Each tablet has 6 mg of drug

Total drug given to cat

Drug = 3\frac{1}{2} * 6mg

Drug = 21 mg

Now dose given to cat in mg/kg form

Dose = \frac{mass of drug in mg}{mass of cat in kg}

Dose = \frac{21mg}{4.54kg}

Dose = 4.63mg/kg

4 0
4 years ago
Other questions:
  • When i stand halfway between two speakers, with one on my left and one on my right, a musical note from the speakers gives me co
    8·2 answers
  • Which of the following is an example of a rotating body?
    9·2 answers
  • A child slides down a frictionless playground slide from a height of 2.1 m above the ground. If she starts with an initial speed
    12·1 answer
  • Which of the following pressures rises and falls with the phases of breathing but eventually equalizes with the pressure of the
    8·1 answer
  • What is universe made off​
    9·2 answers
  • Water waves and light waves are both examples to
    11·1 answer
  • Could there ever be a situation where a small sports car could have more inertia than a big bus?​
    7·1 answer
  • Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going
    15·1 answer
  • A solid sphere is placed on a frictionless floor in a very long corridor and is given a quick push so that it begins to slide, w
    6·1 answer
  • What data will the simulation provide about your design? You will also need a control (something you don’t change) that you can
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!