Problem Solving Strategies

Yashas

1:09 AM

I have forgot what I had tagged u for lol

Koolman

2:19 AM

@blue what are other questions

in mains

Madhuchhanda Mandal

5:01 AM

"
the smaller slit gives out light on the other side with lower power
from the lens perspective, the intensity of light is smaller
P is proprotional to I
becaz at the lens surface area
u do P source/total area to get intensity
the total area is the same for both soruces at the lens
let me upload a pic" @Yashas Are you sure that it's the reason?

Yashas

check the pic

Madhuchhanda Mandal

Checked the Pic

Yashas

what's wrong with the pic?

I wasn't able to explain it in words so I used a pic

there are several ways you can interpret my words but the pic tells what I meant

Madhuchhanda Mandal

You mean something like this:

As the area of the lens on which light from both the slit fall is constant, P proportional to I

Right?

Yashas

not the area of the lens

" total area is the same for both soruces at the lens " = if you draw a virtual sphere surrounding the power source, you get the same area for both

as d << D

Madhuchhanda Mandal

5:05 AM

"Total area"?

The area on which light from the slits fall. Isn't it?

Yashas

yes

in that problem, it is a hemisphere

Madhuchhanda Mandal

You mean the half of the sphere ahead of the slit?

Yashas

yes

Madhuchhanda Mandal

First of all if we consider hemisphere there ends the whole story

Yashas

why?

Madhuchhanda Mandal

5:14 AM

Because the lens covers only a part of the hemisphere and hence in the final expression P will be only a fraction of P

Yashas

It is proportional! The constant cancels out everywhere.

Madhuchhanda Mandal

(See the last equation of the original problem posted)

I think you will understand what I am saying

Koolman

which question you are discussing

Madhuchhanda Mandal

Yashas

AHAHAHA

I see it now

the point D is not at the same location

Koolman

5:19 AM

so whats the confusion in it

Yashas

wait

I am confused now

Madhuchhanda Mandal

I am not

Yashas

the formula tells u are correct

but shudn't u get all the power from A and B which hit the lens at D

oh nvm

Madhuchhanda Mandal

The whole things boils down to : no diffraction

There lies the key I think

Yashas

?

Madhuchhanda Mandal

5:21 AM

Cylindrical wavefront will be transmitted as cylindrical wavefront only not spherical wavefront(cause it is said to be aperture not slit and r>>>lambda, hence no diffraction)

Yashas

what's difference :D

b/w aperture and a tiny hole

Madhuchhanda Mandal

Diffrence is that slit : r->lamda . Aparture r>>lambda

Yashas

I think the formula is correct

I did not understand

Madhuchhanda Mandal

The formulae is absolutely correct

Yashas

o0

Madhuchhanda Mandal

5:24 AM

The reason is not

Yashas

let me make the correction

the raidus of the hemisphere must be the distance b/w the slits and the poitn D

I earlier told lens

now A cancles out everywhere

the lens effectively makes the interference happen at D

Madhuchhanda Mandal

You can't consider sphere at first place

Yashas

why?

Madhuchhanda Mandal

Due to cylindrical wavefront

Yashas

how can a slit produce cylindrical wavefront?

Madhuchhanda Mandal

5:26 AM

Read the question. Its not slit

Yashas

I meant tiny holes

point sources

we consider the slits to be point sources

Madhuchhanda Mandal

Do you know the reason why you consider so?

Yashas

because they are really tiny

Madhuchhanda Mandal

Due to diffraction. Here no diffraction will take place as r>>lambda.

Yashas

I am so confused now

if the slits had finite gap

Madhuchhanda Mandal

5:28 AM

Slit never produce wavefront

Yashas

then there would be diffraction

but the slits are very tiny

so we can ignore it

can consider spherical wavefronts from the slits

Madhuchhanda Mandal

0.001 slit cannot diffract 0.0000001 lambda

See the question

Yashas

AHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

circular apertures with finite radii

hmm

Madhuchhanda Mandal

That's what I am trying to say since last 10min

Yashas

how does it affect my argument though?

@blue This discussion is abt ur question.

Madhuchhanda Mandal

5:31 AM

It does not affect. This changes the argument itself

Anonymous

@Yashas What's going on. Tell me in short!

Yashas

I am confused

the discussion started because you claimed that my reasoning was wrong, right?

Madhuchhanda Mandal

In short : We are re attempting the question

Absolutely

Yashas

@blue I don't know. I am confused.

I am totally confused.

@MadhuchhandaMandal What exactly is wrong with my argument?

Anonymous

There won't be diffraction.

Anonymous

5:33 AM

What's confusing about that?

Madhuchhanda Mandal

I) Spherical wavefront II) Even if we consider spherical wavefront, you cannot prove the formulae to be true by cancellation of constants

Yashas

@blue @MadhuchhandaMandal claims that the explaination I gave you yesterday for that power formula is wrong.

Madhuchhanda Mandal

Of course

Yashas

What is it not spherical wavefront?

it is relatively really really tiny

oh my

you won't have spherical wavefront

the wavelength is 200nm

the slit is 1mm

you would get a straight stream of light

Madhuchhanda Mandal

Ok lets consider for a moment spherical wavefront. Now try to explain

Yashas

5:35 AM

with little diffraction at the edges'

Anonymous

There will be multiple point sources

Yashas

ok now it kills my argument

Anonymous

But that doesn't change the basic arguments

Yashas

yea it doesn't change the argument at the core

the idea is the same

it is just that it is not a sphere

Anonymous

You can completely explain interference with multiple diffraction

Yashas

5:36 AM

the energy is going to be distributed uniformly

Madhuchhanda Mandal

Try to explain

Yashas

@MadhuchhandaMandal ?

spherical wavefront is wrong

Anonymous

@Yashas Take any two points in each slit. It will be a sphere for each point.

Anonymous

Doesn't matter how many spheres are formed

Yashas

@blue I used the sphere for the whole thing.

Anonymous

5:37 AM

@Yashas That.....is wrong....

Yashas

yea that is wrong

Madhuchhanda Mandal

OMG!!! If the energy is distributed uniformly then how the hell are you going to get at D power received = Transmitted Power through slit ?

Yashas

and for points close to each other

but for r << D

the energy is distrubuted uniformly

Madhuchhanda Mandal

THE ENERGY IS NEVER DISTRIBUTED

Yashas

yes I know now

the light beam goes straight through

it does not form a spherical wavefront

it diffracts just a little at the edge

but we ignore it

all of them hit the lens

so energy is not distrubited at all

Madhuchhanda Mandal

5:39 AM

Yes

Yashas

and they interfere at point D

I mixed up sentences

which caused the latest confusion :P

I got punished for not reading the whole question :D

Madhuchhanda Mandal

Yeah... Of course

Now you have just missed a vital mechanics

Yashas

?

Madhuchhanda Mandal

How do you prove the power formulae?

Yashas

my previous reasoning is false for this problem

but for spherical wavefronts, it is correct

for this case, u use area of the slit

Madhuchhanda Mandal

5:43 AM

And now?

Yashas

the same reasoning but instead of a sphere, u need to use the area of the slit to get intensity

the area once again cancels out

Madhuchhanda Mandal

Slit A will be 4P if Slit B will be P

No they will never cancel out

Yashas

aw the areas are not same

got punished for the second time

for not reading the whole question

Madhuchhanda Mandal

So?

How do you explain it?

Yashas

area of the point of intersection?

infinitesimally small area

Madhuchhanda Mandal

5:48 AM

Means?

Yashas

let the point D have finite area A

you get Ia and Ib to be Pa/A and Pb/A

Madhuchhanda Mandal

Yeah you are close

Just few steps

Yashas

A cancles out? :/

take the final power as P

PA gives the final intensity

WAIIT WHAT THE HELL

WE DONT LOSE POWER DUE TO INTERFERENCE

Do we?

Anonymous

The spherical wavefront argument still holds. $$I_{A}=\frac{I_1A_1}{\pi R^2}$$

Yashas

after they cross point D, they will have Pa and Pb

again

Madhuchhanda Mandal

5:51 AM

We don't loose power. We redistribute it.

Anonymous

There will be coherent waves from both the slits

Yashas

what is r @blue

Anonymous

@Yashas Radius of the wavefront which strike at point D

Madhuchhanda Mandal

r????? Where is it?

Anonymous

$R$ is same for both the slits

Yashas

5:53 AM

Anonymous

You don't need to know its value

Anonymous

It will cancel out automatically

Anonymous

@Yashas WTF....

Madhuchhanda Mandal

What's that?

Yashas

the slit widths are huge

There is little diffraction at the edges

otherwise, it is a straight beam

Madhuchhanda Mandal

5:54 AM

Ignore diffraction please

Anonymous

@Yashas LOL...if slit width is that huge their won't be visible interference patterns at all....

Yashas

I exaggerated the diffraction at the edges at the pictures though

@blue yes there won't be any at the slits

Madhuchhanda Mandal

Why the hell ?@blue

Yashas

but there is a refractive material placed in the path

Anonymous

Just take spherical wavefronts

Yashas

5:54 AM

it causes a phase difference

Anonymous

Let me show you a pic

Yashas

when the lens bends the two beams, u get interference

Anonymous

@MadhuchhandaMandal Because a path difference will be introduced and they won't be coherent...multiple interference will occur

Madhuchhanda Mandal

Absolutely

@blue non-coherency and path difference are completely different

Anonymous

@MadhuchhandaMandal nope.

Madhuchhanda Mandal

5:56 AM

Yup

Yashas

I did not show the refractive material explicitly in my diagram

but I meant that was going to happen

otherwise, the P would be Pa + Pb at the focus

Anonymous

Draw it like this ^

Anonymous

Forget the convex lens...that doesn't matter (it just focuses the rays)

Madhuchhanda Mandal

The diagram is making me go mad lol

Anonymous

5:58 AM

Coherence (physics)

In physics, two wave sources are perfectly coherent if they have a constant phase difference and the same frequency. Coherence is an ideal property of waves that enables stationary (i.e. temporally and spatially constant) interference. It contains several distinct concepts, which are limiting cases that never quite occur in reality but allow an understanding of the physics of waves, and has become a very important concept in quantum physics. More generally, coherence describes all properties of the correlation between physical quantities of a single wave, or between several waves or wave packets...

Anonymous

two wave sources are perfectly coherent if they have a constant phase difference

Madhuchhanda Mandal

@blue Looooolllllllll .... Read it once

Anonymous

I don't need to

Yashas

@blue me and @MadhuchhandaMandal are actually ignoring the diffraction at the edges

u won't get the spherical wavefront u have drawn @blue

Madhuchhanda Mandal

Phase difference and path difference are different. Right ? @blue

Anonymous

6:00 AM

@Yashas I haven't drawn it. The diagram is from a standard source.

Anonymous

@MadhuchhandaMandal They imply each other.

Anonymous

Path difference implies phase difference and vice versa.

Madhuchhanda Mandal

@blue LOL

Anonymous

lol

Anonymous

^

Madhuchhanda Mandal

6:01 AM

@Yashas what do you say?? Path difference means non-coherency?

Yashas

no

it depends on the value

Madhuchhanda Mandal

Make @blue Understand

Yashas

$\Delta \phi = k\delta x$

Anonymous

^

Yashas

wait

coherency ?

I think that has nothing to do

that has to do with the source

Madhuchhanda Mandal

6:03 AM

Absolutely

Yashas

I think @blue was talking about the phase difference caused by the path difference

not abt the phase difference b/w the two sources

Anonymous

Yep ^

Madhuchhanda Mandal

Of course. Does that destroy coherency?

Never EVER

Anonymous

If slit is too wide too many interferences will take place

Anonymous

That's why you will never observe interference will a torch and two slits

Anonymous

6:04 AM

Try it at your home

Yashas

@blue never talked about coherence

Madhuchhanda Mandal

Read 10 messages up

Yashas

4 mins ago, by blue

Path difference implies phase difference and vice versa.

Anonymous

Well, you misinterpreted what I meant. Move on.

Yashas

he did not mean the phase difference b/w the sources there

Madhuchhanda Mandal

6:05 AM

5 messages up of that

Because a path difference will be introduced and they won't be coherent

@blue said that

Anonymous

I did.

Anonymous

I stand by it.

Anonymous

Try the experiment at your home with large slits.

Anonymous

You will never get interference fringes.

Anonymous

The slit sizes must be comparable to the wavelength.

Madhuchhanda Mandal

6:08 AM

Because the sources won't be coherent... The path difference introduced by material has absolutely nothing to do with it

Anonymous

I never said the sources won't be coherent -_-

Madhuchhanda Mandal

Whoa!!!

Good Day... Need to study now. Bye

Anonymous

k.

Anonymous

bye.

Koolman

The velocity of cylinder due to left wedge is $-v/4 \hat i - \sqrt3 v/4 \hat j$
Velocity of cylinder due to right wedge is $2u \hat i$
So net velocity is $\sqrt{13}/2$

What mistake I have done

@Yashas now I think it has calm.down

Yashas

6:16 AM

hmm

leme see

pff

I read the solution

thinking it was part of the problem

Koolman

lol

Yashas

I can't make sense out of the solution though

Anonymous

Take point of contact. The velocity will be same along common normal.

Yashas

as the ball goes down, it would apply force on the two wedges

it would cause an acceleration on the wedges

they took the velocity as constant

ohh that must be the constraint

the wedges are forced to go at those speeds

Koolman

Anonymous

6:20 AM

The solution is okay!

Koolman

@blue direction normal to PQ

Yashas

yea the solution is ok

@Koolman Sit on the wedge P

Anonymous

Take point of contact of cylinder with the right cylinder

Yashas

take that as the frame of reference

the other wedge is moving at 3u away from u

the cylinder is going down with the speed v

the horizontal component of v must be equal to 3u

becaz the cylinder stays in contact with the right wedge always

lately, I have started misreading questions too much

skill patrol

@Koolman I can think of a way ;-)

To get around the ban.

Koolman

6:23 AM

What @skillpatrol

@Yashas Thats the solution given

Yashas

yea it is correct

Koolman

But I want to know mistake in my method

Yashas

I can't understand what you'ev done

Koolman

I have simply applied constrained due to both wedge

Yashas

can you explain what you have done?

We have 7 members online in this room simultaneously. We hit a new record I guess.

skull petrol

6:28 AM

6 actually

Koolman

The velocity of cylinder due to left wedge is $-v/4 \hat i - \sqrt3 v/4 \hat j$

As velocity if cylinder and left wedge is same in direction perpendicular to PQ

skill patrol

You counted me twice ;-)

Koolman

constrain @Yashas

Yashas

@skillpatrol There is a hidden user @Fl.pf.

hover the cursor after MadhuchhandaMandal

skill patrol

I see.

Koolman

6:29 AM

@blue you mean right wedge

Yashas

@Koolman I cannot understand. What you have done does not make sense.

Koolman

@Yashas He is always online

Yashas

and we never know lol

Anonymous

@Koolman yeah

Koolman

@Yashas What you not able to understand

Yashas

6:31 AM

The velocity of cylinder due to left wedge is $-v/4 \hat i - \sqrt3 v/4 \hat j$

how did u end up with that?

Anonymous

Okay, see the cylinder is constrained to move only along the incline

Anonymous

So you can take the component of either of the wedge along the common normal

Anonymous

And equate it to the component of the cylinder's velocity in that direction.

Anonymous

Take common point with either right wedge or with left wedge

Koolman

yes that what I have done

skill patrol

6:35 AM

@Koolman I didn't want to mention opening a second account in front of the mod. But if the two accounts do not interact it is allowed.

Koolman

velocity of cylinder normal to PQ = $u\sin 30 $

now resolve it into x and y direction

@skillpatrol That would be nice idea

Thanks

This how I got that @Yashas

Anonymous

Yashas

@Koolman There shud be zero relative motion along the perpendicular line. You shud equate the cylinder's normal component with the wedges' normal component.

^ that is method I

as blue showed

the method II is what is given in ur solution book

Koolman

@Yashas What is cylinders normal

Yashas

6:40 AM

I meant the direction of the normal force b/w the cylinder and the wedge

I am bad at English :p

Koolman

Now I am more confused

The velocity of cylinder is along left wedge