Problem Solving Strategies

harambe

5:23 AM

@JohnRennie good morning. You free

John Rennie

@harambe morning :-)

I'm just going to make a coffee then I need about 15 minutes to do some work.

harambe

Okay. See you around an hour

ayc

@JohnRennie morning!..Are you free?

John Rennie

@ayc in about 15 minutes ...

ayc

@JohnRennie I'll be waiting ..Just ping me

ayc

5:58 AM

@JohnRennie This is my question:we know that magnetic field does no work on a charge moving with constant velocity.So the speed of the particle remains constant.But its velocity vector changes.That means it is accelerating.In this case,since the charge is undergoing an acceleration would it emit EMW?

John Rennie

@ayc yes. This type of radiation is known as Bremsstrahlung or synchrotron radiation. For example accelerators like the LHC emit this type of radiation when they use magnetic fields to make particles go in a circle.

ayc

@JohnRennie Oh yeah.We talked about it yesterday.I forgot

John Rennie

@ayc actually I had forgotten as well :-)

ayc

@JohnRennie See you with a new doubt.Thank you !

harambe

6:19 AM

@JohnRennie hi. Are you free now

John Rennie

@harambe yes

harambe

imgur.com/a/2c1Hdam

Q34

I set mv^2/4=10.2eV

I converted every to joules and tried to get the answer

But it's not matching.. Is this correct

John Rennie

This is basically the same as that neutron-H atom problem we looked at yesterday.

The ionisation energy is 13.6eV, so in the COM frame the collision energy has to be 13.6eV i.e. the KE of the atoms has to be 6.8eV.

OK so far?

harambe

Wait I was using excitation energy instead of iionisation energy

I get my mistake

@JohnRennie can't we use total kinetic energy of the atoms to be 13.6 ev

And calculate velocity from there

John Rennie

No. Can you guess why not?

harambe

6:31 AM

Some energy is used in excitation energy

John Rennie

The collision must conserve momentum.

harambe

Okay

John Rennie

That means atom A cannot come in with energy 13.6eV, ionise atom B and come to a halt. That would not conserve momentum

The requirement that momentum be conserved places a constraint on how much of the incoming atom's energy can be used to ionise the other atom.

This is why it is best to work in the COM frame then transform back to the lab frame for the final result.

In fact the problem is trivially easy if you do it that way ...

harambe

@JohnRennie So I write the total kinetic energy is coming frame

And then equate it to 13.6 ev

John Rennie

is coming is your phone autocorrect of in COM ?

harambe

6:35 AM

Yes....

COM frame

John Rennie

Look at it this way ...

If the incoming atom has a speed $v$ then in the COM frame the two atoms have speeds $+v/2$ and $-v/2$. Yes?

harambe

Yes

John Rennie

And KE is proportional to v^2 so in the COM frame the KE of the incoming particle is 1/4 of its value in the lab frame.

harambe

Yes

John Rennie

Now, in the COM frame the total momentum is zero, so the two atoms can come in with equal and opposite speeds, use all their KE to ionise one of the atoms and come to a stop. This conserves momentum so it is allowed. So in the COM frame the minimum total KE is 13.6eV.

harambe

6:39 AM

Yea

So I calculate velocity now and convert it into lab frame

John Rennie

That means both atoms have a KE of 6.8eV, and we've just worked out that in the lab frame the KE of the incoming atom is four times greater i.e. 27.2eV.

harambe

Ok

John Rennie

So you can just use $\tfrac{1}{2}mv^2 = 27.2eV$ in the lab frame.

harambe

Got it

John Rennie

Not that only half the KE can be used to ionise the atom. That's the conservation of momentum imposing a constraint on the collision.

harambe

6:45 AM

How did you conclude half

John Rennie

The initial KE is 27.2eV. Half of this, 13.6eV, is used to ionise the atom, and the remaining 13.6eV goes into the KE of the atoms after the collision.

harambe

@JohnRennie Finally I also got the answer

John Rennie

@harambe cool :-)

harambe

@JohnRennie in inelastic collision, only momentum is conserved and we can't use energy equation right

John Rennie

@harambe correct

Well, energy is still conserved but you have to count all the energy including the energy that goes into exciting the atoms or whatever.

So you can still use the energy equation provided you count all the energy. Just counting the KE won't work.

harambe

6:54 AM

Okay

@JohnRennie Quoting from wiki

In nuclear physics, an inelastic collision is one in which the incoming particle causes the nucleus it strikes to become excited or to break up

Does that mean all inelastic collisions in nuclear physics, the kinetic energy is spent in exciting the atoms

John Rennie

Energy is always conserved, so if energy "disappears" in a collision then that energy must be going somewhere. The question is where can the energy be going?

In the case of something colliding with a nucleus the only place energy can be going is into exciting the nucleus.

harambe

What about ionising

John Rennie

You don't ionise nuclei ...

harambe

Oh lol

Got it

John Rennie

Don't confuse atomic collisions, like the problem we've just done, with nuclear collisions. Excitations of nuclei happen at energies of many MeV while for atoms it's at most a few keV for the innermost electrons in heavy atoms.

harambe

7:10 AM

@JohnRennie the atomic collisions we have discussed till now we're all inelastic, right

John Rennie

@harambe yes. An elastic collision is one in which the total final KE is equal to the total initial KE. If some KE gets used up in exciting or ionising atoms then by definition that collision is inelastic.

harambe

@JohnRennie imgur.com/a/2c1Hdam

Q35

The collision is inelastic when kinetic energy is spent in exciting

The excitation energy is 10.2 eV

And the frame is COM too

John Rennie

Yes, so the question is very easy.

harambe

So the atoms can stop and it will not affect the momentum

Got it

@JohnRennie got the answer but why have they written the collision may be inelastic or perfect

Can the collisions vary between perfect or non perfect

John Rennie

A partially inelastic collision is one in which some but not all of the KE is used to excite the atom. A completely inelastic collision is one in which all of the KE is used in exciting the stom.

harambe

7:44 AM

@JohnRennie in. Photo electric experiment, the electrons that are ejected due to incident light gain kinetic energy because of incident light. Does this mean all electrons near the surface come out with kinetic energy equal to the energy provided by incident light on metal or can the electrons still lose energy even if itnis near the surface

John Rennie

The PE experiment is more complicated than it appears at first sight.

When a photon produces a photoelectron that electron is travelling in the same direction as the incident photon i.e. downwards into the metal.

harambe

Okay

John Rennie

For a photoelectron to be ejected from the surface the initial electron has to collide with something and rebound back towards the surface. Or it might transfer its energy to a different electron that then escapes the surface.

This is a rather improbable event, and in fact the vast majority of the initial photoelectrons never escape. Only about one in a million, or at best 1 in 100,000, photons actually produces a photoelectron from the surface.

harambe

Okay but what about the free electron of the metal

They can gain energy from photon collision.. I read this in my textbook

John Rennie

By free electrons I assume you mean the electrons in the conduction band, and yes for visible light all the photoelectrons come from the conduction electrons.

But my previous statement still holds. When a photon transfers energy to a conduction electron conservation of momentum means that conduction electron starts moving in the same direction as the photon.

harambe

7:53 AM

Okay

John Rennie

Anyhow, the collisions involved in making the electron bounce back towards the surface are partially elastic to some extent. That means some or all of the energy of the photon can be lost in those collisions.

The result is that the photoelectrons that come from the surface have a range of kinetic energies from a maximum of $h\nu - \phi$ right down to zero.

harambe

What does this $h\nu - \phi$ represent here

John Rennie

$h\nu$ is the energy of the incident photon and $\phi$ is the work function of the metal.

harambe

Okay

@JohnRennie Why is The frequency of the photon with energy equal to work function is called threshold frequency.

John Rennie

If we take the PE outside the metal to be zero (as usual) then the topmost electrons in the metal lie at an energy of $-\phi$. That is is always costs an energy $\phi$ to remove an electron even if we don't give the electron any KE at all. You can think of $\phi$ as being the ionisation energy of the metal.

harambe

8:03 AM

Okay

John Rennie

That means if $h\nu < \phi$ the photon cannot eject an electron, just like photons with an energy less than 13.6eV cannot ionise a hydrogen atom.

harambe

@JohnRennie that would explain my previous question too

Okay

John Rennie

Ah, sorry, I didn't realise you hadn't done the work function yet.

harambe

@JohnRennie even if energy of photon is greater than work function, what if by the time it reaches the surface electron, it's energy gets reduced below work function due to succesive collisions with other electrons

John Rennie

@harambe I think, but don't know for sure, that the most probably scattering event for the photon is to transfer all its energy to an electron in one go. So as a general rule photons don't get partially scattered and lose a bit of energy before their final collision.

harambe

8:10 AM

Okay. Got it

@JohnRennie files.mtstatic.com/site_4334/125522/…

In the graphic of current vs frequency, why are they independent

Greater frequency means more electrons will come from surface of metal

With more energy of course

John Rennie

Remember that as long as the frequency is above the threshold one photon = one photoelectron. If you change the frequency but keep the number of photons constant then the number of photoelectrons will be constant.

NB in most PE experiments the current just measures the number of photoelectrons per second. The current is not dependent on the photoelectron energy.

harambe

Okay got it

John Rennie

That graph is actually rather misleading.

If you keep the intensity constant then the number of photons per second decreases as we increase the frequency. This is simply because each photon has more energy, $hf$, so for a fixed energy higher $f$ means fewer photons.

harambe

8:26 AM

@JohnRennie didn't get it

If frequentcy is increased then photos will have more energy. How will this relate to intensity which depends on number of photons

John Rennie

Suppose we keep the light intensity constant as we change the frequency. The intensity is power per unit area i.e. joules per second per square metre.

So the number of photons per second per square metre is the intensity divided by the photon energy $hf$.

harambe

Oh okay

John Rennie

That's why graph C is a bit misleading. The obvious way to do the experiment would be to keep the intensity constant as you increase the frequency. If you do that then the current will be proportional to 1/f.

The graph must be keeping the number of photons per second per unit area constant i.e. it is increasing the intensity as the frequency is increased.

harambe

9:38 AM

@JohnRennie absorbing photon is a positive work or negative

I think positive because energy is increasing

John Rennie

I'm not it's really work in the usual sense of the word.

harambe

Okay

harambe

9:52 AM

@JohnRennie imgur.com/a/mPycWDv

Q4

I found the energy of the photon

Then I divided it by the power

Didn't get the answer.. Am I missing something

John Rennie

Well, you divide the power by the photon energy not the other way around. I assume thats what you meant.

What did you take as the power?

harambe

Yea I divided power by energy

10 W

John Rennie

harambe

So I should do 60 percent of energy or power

John Rennie

The lamp consumes 10W of electricity, and it is 60% efficient, so the power of the light produced is 10 x 0.6 = 6W.

harambe

9:58 AM

Okay so the photon is the energy emmited front this energy?

Electric energy I mean

John Rennie

Yes. Divide 6W by the photon energy to get the number of photons per second.

harambe

Got the answer

John Rennie

Cool :-)

harambe

How to calculate for cubic metere

John Rennie

The first step is to calculate the number of photons emitted per square metre per second at the Earth's surface.

harambe

10:11 AM

Intensity/Energy

John Rennie

Does that say Earth's surface? You managed to get the word PRO right over it.

harambe

Yes

It says earth

John Rennie

OK, so you calculated the number of photons/sec/m^2 in part (a). Yes?

harambe

Yes

John Rennie

In one second a photon travels $c$ metres, so in one second the contents of a volume 1m x 1m x c metres of photons hit every square metre on the Earth. Yes?

harambe

10:15 AM

Yes

@JohnRennie on second thought I can't visualise it

John Rennie

Consider a box 1m x 1m x c metres standing on the Earth. Every photon inside that box will hit the Earth in the next second.

harambe

Okay

John Rennie

And only the photons inside that box can hit the Earth in the next second, because any photon farther away than $c$ metres will take more than a second to reach the Earth.

harambe

Okay

John Rennie

So all the photons that hit a square metre of the Earth in on second fill a box of volume 1 x 1 x c.

harambe

10:21 AM

Okay

John Rennie

In part (a) you calculated the number of photons per second per square metre hitting the Earth. Call the number $N$. All these $N$ photons are contained in a box of volume 1 x 1 x $c$, so the number density (number of photons per cubic metre) is $N/c$.

harambe

Got it

@JohnRennie doubt

John Rennie

@harambe yes?

harambe

Sun light from sun takes roughly 8 minutes to reach earth. How would photons travel there in 1 sec

With speed c

John Rennie

I wasn't counting all the photons in transit between the Sun and the Earth. I was only counting all the photons in a box of height 1 light second above the Earth.

harambe

10:34 AM

Okay got it

harambe

10:49 AM

@JohnRennie (c)

I multiplied intensity by area of earth

Then divieded that by energy of photon

Answers not matching

John Rennie

The question asks how many photons per second the Sun emits, not how many photons per second hit the Earth.

harambe

Oh dang

John Rennie

I sure it's not really necessary to say, but it is important to read the question carefully. In the exam you don't have someone like me around to correct you.

harambe

Yes you are right.

@JohnRennie any idea how to solve this part then

I can't seem to think one

John Rennie

11:05 AM

@harambe you've already calculated the number of photons per square metre per second at the Sun Earth distance. Yes?

harambe

Yes

John Rennie

How many square metres are there at the Sun-Earth distance?

(it's not a trick question)

@harambe hello?

harambe

Thinking

John Rennie

@harambe Call the Sun-Earth distance $r$. What is the surface that is everywhere a constant distance $r$ from the Sun?

harambe

Eh?

Am I supposed to say earth's surface

John Rennie

11:12 AM

We are asked to find the total number of photons per second coming from the Sun. So if we consider a surface around the Sun with the Sun at the centre we just just have to work out how many photons per second pass through this surface.

OK so far?

harambe

Ok....

John Rennie

And the total number of photons per second passing through the surface will be the number of photons per square metre at the surface times the area of the surface.

harambe

Yes

John Rennie

Like that, where I've just drawn some random surface with the Sun in the middle.

(the Sun is the yellow circle)

But calculating the area and the photon intensity for a random surface like that is going to complicated. It would be far easier if it was a more symmetrical surface ...

harambe

Like sphere

John Rennie

11:19 AM

Yes, like a sphere. Maybe a sphere with a radius equal to the Sun-Earth distance since we have already calculated the photon intensity at the Sun-Earth distance ...

harambe

Okay

harambe

11:35 AM

@JohnRennie got it

John Rennie

@harambe Cool :-) The idea is simple when you've been introduced to it. The same technique is used extensively in electrostatics where we call the surface a Gaussian surface.

harambe

Yeah

harambe

12:34 PM

@JohnRennie imgur.com/a/iGAfJpE

Q20

Can you explain to me what does does electric field do here

John Rennie

Light, or indeed any EM wave, is an oscillating electric field. So when the question talks about the electric field going to zero this is just a complicated way of telling you the frequency of the light.

harambe

@JohnRennie de broglie wavelength is associated with particle nature or light behaves as particles

And einstein gave wave nature, right

John Rennie

I'm not sure what you are asking

harambe

Leave that xd.I was actually trying to ask something weird

John Rennie

If you're wondering about wave particle duality then it's a surprisingly simple idea but it requires you to have a basic grasp of quantum field theory.

harambe

12:42 PM

@JohnRennie quantum field theory is in the course major physics students, right

John Rennie

@harambe it's usually an option that the mathematically inclined students might want to in their final year of a degree. QFT is really hard to understand rigorously. The maths involved is super scary. However it's not too hard to get a rough idea of what is going on.

harambe

@JohnRennie is schrondinger equation also solved in QFT

I remember my teacher saying the math required 1-2 day solving

John Rennie

No, the Schrodinger equation is non-relativistic quantum mechanics and it's actually not that hard. Most JEE students will have enough maths to be able to handle it.

Quantum field theory is a whole new level of complexity!

If you did physics at university you would study the Schrodinger equation as a first year course.

Sid

@harambe You will study Schrodinger's equation in first semester

(Of any University)

harambe

@Sid okay. I wonder why my teacher said that then

Sid

12:53 PM

@harambe They probably didn't want to consume valuable time explaining you guys about it.

John Rennie

@harambe even basic QM involves some ideas that are hard to get your head round. The maths isn't that hard but the concepts are.

Actually the same is true of special relativity. The maths is just basic calculus and JEE students will know this. But the ideas involved are very ... erm ... odd! :-)

Sid

(You will also read about Special relativity in first year of any Uni)

harambe

Wow. There is a chapter on special relativity in hcv too

John Rennie

Beware of "one chapter" introductions to special relativity as they tend to oversimplify.

harambe

@JohnRennie got it

ayc

1:40 PM

anyone?

Dante

1:59 PM

Hello, anyone? @JohnRennie or @sammygerbil

Dante

2:11 PM

First doubt, when Pn junction diode is reverse biased, the voltage is along the direction of electric field of the depletion region. Then why does the 'voltage drop across the depletion region'?

This is what is given in my book.

Dante

3:49 PM

@Blue Hi, are you free?

Anonymous

4:01 PM

6. Physics Lecture of 12th Class| PN Junction & Sem Diodes | Reverse Biasing | by Ashish Arora

►

Anonymous

Does this help?

Anonymous

@Dante Unfortunately not. If you have any specific question, ping me with it. I'll try to answer it by tomorrow.

Dante

Ok, unfortunately I don't have any specific question.

But thanks for the video!

Ice Inkberry

4:28 PM

I have a small doubt in errors and measurements. The statement says:

For example, in the experiment on finding the focal length of a convex lens, the object lens(u) is found by subtracting the positions of the object needle and the lens. If the optical bench has a least count of 1 mm, the error in each position will be 0.5 mm. So, the error in the value of u will be 1 mm. The doubt I have is, how is error in each position 0.5 mm? Isn't the error directly the least count(unless otherwise mentioned). So, 1 mm + 1 mm = 2 mm for (u).

Abcd

@sammygerbil Please ping me when you are here for problems.

@Dante its given properly in RnH too.

Dante

Hmm, I don't have unfortunately :/

Abcd

@Dante its given in HCV to

Dante

I see, I read it from NCERT, not clear enough, will read it in HCV now.

Esha Manideep

5:40 PM

@JohnRennie hi!

John Rennie

@EshaManideep hi

Esha Manideep

I have this question, I have a laser light run by some solar cell, but I now I focus the parallel beam onto a parabolic reflector which converges the light rays to a very small point, and now I can put an object at that focus and heat it to whatever temperature I can by reducing the area, ultimately greater than that of sun, the against SLT.

What is wrong with my argument??

sammy gerbil

9

A: Is it possible to focus the radiation from a black body to make something hotter than that black body?

If you use complicated routes redefining "focusing" towards generating the temperature, yes. Physicists at CERN's Large Hadron Collider have broken a record by achieving the hottest man-made temperatures ever - 100,000 times hotter than the interior of the Sun. Scientists there collided...

John Rennie

As your object heats up it starts emitting light. The same lenses that focussed the light from the Sun (or whatever) send the light from the object back to the Sun. The end result is that the fraction of light from the Sun that goes into heating the object decreases as the object gets hotter.

sammy gerbil

@Abcd Available now (but going out for about 30 minutes in 1 hour.)

Esha Manideep

5:51 PM

@JohnRennie Alright, do u remember a question I asked u few days ago about focusing sun rays through a lens....

John Rennie

@EshaManideep sorry, no.

Abcd

@sammygerbil A parallel beam of sodium light of wavelength $5890 $ A is incident on a thin glass plate of $\mu = 1.5$ such that angle of refraction is $60^\circ$. The smallest thickness of the plate which will make it dark by reflection is?

@JohnRennie Do you know about resolving power of optical instruments?

Esha Manideep

@JohnRennie Alright let me send the question again.

John Rennie

@Abcd possibly, it depends on exactly what you want to know.

Abcd

@JohnRennie i will ask in morning coz you are going rn probably?

John Rennie

5:55 PM

Yes, I'm going soon. I'll be around as usual tomorrow.

Esha Manideep

Sun rays are focused with a lens of diameter d and focal length f to the black side of a thin plate. The one side of the plate is perfectly black and the other side is perfectly white. Angular diameter of the sun is aloha and it's intensity on the surface of the earth is I. I using thermodynamic arguments find the maximal focal length-to-diameter ratio of a lens.

Abcd

@JohnRennie i cant understand my textbook, there are no good youtube videos on it. Thats why i need to understand the concepts and stuff. (tomorrow)

@sammygerbil The unusual part of the problem is that in all problems I have done so far, incidence and refraction, reflection was normal

@sammygerbil Given the $\mu$, Isn't 60 degree a very large angle of refraction?

Esha Manideep

I did this one by finding the image diameter and equating the images area into t^4 to the power falling on it and the temperature of it equalling the sun

John Rennie

OK ...

Esha Manideep

Wait,...

Nonsense 😂, ignore it. I sorry for wasting your time

John Rennie

6:01 PM

No problem :-)

sammy gerbil

@Abcd Yes it is too big. Maximum angle of refraction should be 41 degrees (=critical angle).

Abcd

@sammygerbil Is the problem wrong?

@sammygerbil when I used snell's law , sin i was coming out to be greater than 1 which aint possible

sammy gerbil

@Abcd Yes. That angle of refraction cannot happen. The question is faulty. Probably not worth continuing with it.

Abcd

@sammygerbil one more

@sammygerbil I tried 20th three times in exam and got answer as $2\sqrt 2$, can you please see what answer you are getting?

sammy gerbil

@Abcd (You could change the angle of refraction to something physically possible, and see if you can solve the problem then. You will not be able to match the answer given, but you can still learn from the problem.)

Abcd

6:09 PM

@sammygerbil let me try 20 again in front of you

See:

sammy gerbil

ok

Abcd

@sammygerbil one of my batchmates claims to have got the answer w/out calculating sin i

@sammygerbil he also says incidince angle isnt needed to solve it

20:

$R = \lambda N$

$N = N_o e^{-kt}$

$\lambda =\dfrac{\ln 2}{t_{1/2}}$

$\implies \lambda_1 = \dfrac{\ln 2}{2}$

sammy gerbil

@Abcd Ask for his solution.

Abcd

$\implies \lambda_2 = \dfrac{\ln 2}{4}$

$R_1 = \lambda_1N_o e^{- \ln 2}$

$R_2 = \lambda_2 N_oe^{-ln 2/ 2} $

$\lambda_1 / \lambda_2 = 2$

$R_1/ R_2 = 2 e^{- \ln 2/ 2}$

$R_1/R_2 = \dfrac{2}{\sqrt 2}$

$R_1/ R_2 = \sqrt 2$

Damn ittttttttttttttttt

Why doesnt it come in the exam hall.

@sammygerbil Problem 17 please

Dont know how to solve that one.

Language of question aint clear.

sammy gerbil

@Abcd Use the Lens-Maker's Equation?

Abcd

6:24 PM

@sammygerbil whats the diagram, language isnt clear to me'

sammy gerbil

Lens has air on one side, water on the other. Rays which are parallel on the air side come to a focus in the water. Where is that focus?

Abcd

@sammygerbil He sent this to me^

Me: Send solution
He: It's hidden somewhere here *laughing emoji* (sic)

@sammygerbil His solution on the bottom right corner, you can see $2\mu x = \lambda$ stuff there.

@sammygerbil I think I can see his mistake, he is considering it normal incidence on the glass surface, isn't he? But the path difference would change for $60^\circ$ refraction angle

sammy gerbil

@Abcd I wonder if he gave the examiner the same solution...

Abcd

@sammygerbil We aren't supposed to get the solutions checked or show them to anyone. We just have to mark the right answer and move ahead in online mode. And bubble the circle in offline - OMR mode... That's why tricks and stuff work and help a lot

@sammygerbil Have I spotted his mistake correctly?

sammy gerbil

@Abcd Don't know, can't read it.

Abcd

6:32 PM

@sammygerbil let me rotate and send

@sammygerbil clear now^ ?

sammy gerbil

@Abcd I can see working for several questions there, but not #17.

Abcd

@sammygerbil let me circle and send lol

sammy gerbil

@Abcd I need to go out now. Back in about 30 mins.

Abcd

@sammygerbil please ping when back

john

6:56 PM

my question got shut down

im not sure why

-2

Q: Orbit velocity satellite question, is this question ill defined

I am attempting to solve this question but I feel that the question may be ill defined. The question is in two images, the first image and second image. I have tried many attemps at this but nothing seems to make sense. The question seems to be talking about entering an orbit starting from the su...

i wasnt asking for an answer, just a direction

the question could be ill defined, thats all i wanted to know. jeez, why was it put on hold

thanks for all the help

sammy gerbil

@Abcd back

Abcd

@sammygerbil I think user John's messages are better suited for h-bar. Please move them there so that his plea can get some attention.

oh no, he has posted the same thing in both rooms

so leave it

sammy gerbil

7:12 PM

@Abcd I think it is suitable for this room.

Abcd

@sammygerbil well, he was asking the reason why his question was put on hold, not a solution to his problem so I thought...

sammy gerbil

7:25 PM

@john Re: why the question was closed. The main site is not a Problem Solving Site. It is intended for conceptual questions only (although algebraic calculation questions sometimes get through if they are interesting). You failed to identify a conceptual difficulty.

@Abcd True, the issue of why the question was put on hold would be better in the h-bar. But I think @john has posted here to get an answer to the question which was closed. ie What (if anything) is wrong with his calculation?

Abcd

@sammygerbil Ok,please see interference question now

sammy gerbil

@Abcd Your benchmate has not used normal incidence in his calculation. The rays which are reflected from the upper and lower interfaces are parallel and interfere destructively if they are in anti-phase. There is a phase difference due to the different optical path lengths, and there is a phase change on reflection from the upper surface (but not the lower).

sammy gerbil

7:46 PM

However, for the same reason that the angle of refraction cannot be 60 degrees, the ray which is reflected from the upper surface cannot emerge from the flat plate - it is incident at greater than critical angle, so it suffers total internal reflection at the upper surface as well as at the lower surface.

Correction : You are right. He has used normal incidence.

Suppose the light is incident at angle $i$ and refracted at angle $r$. Then $\sin i =n\sin r$.

Suppose the thickness is $t$ and the distance between points at which one ray enters and emerges from the plate (after reflection from the bottom face) is $d$. Then $2t=x\cos r$ and $d=x\sin r$ where $x$ is the path length of the ray in the glass plate.

Abcd

8:07 PM

@sammygerbil he says that he has not used normal incidence

@sammygerbil he says that x is slant length only

sammy gerbil

@Abcd ok. I don't follow what he has done.

Abcd

@sammygerbil he has an argument for angle of refraction too

@sammygerbil he says that medium 1 is not specified.

but we took it as $\mu_1 = 1$

Abcd

9:35 PM

@sammygerbil still here?

@sammygerbil The amplitude of electric field of an EM wave travelling along z axis is 2 V /m. The average magnetic field is:
A) 13.29 * 10^-12
B) 8.86 * 10 ^-12
C) 17.72 * 10^-12
D) 4.43 * 10^{-12}

@sammygerbil am getting C using $\dfrac{1}{2}\epsilon_o E^2$

But Answer is B

sammy gerbil

@Abcd A factor of 2 difference...

How are you using $\frac12 \epsilon_0 E^2$?

Abcd

@sammygerbil energy density of magnetic field = energy density of electric field in EM wave

sammy gerbil

@Abcd So what relation have you got between $E$ and $B$?

Abcd

@sammygerbil $E = Bc$

@sammygerbil aint that relation useless here, why do you need it