@JohnRennie Morning :-)

@Jasmine hi :-)

Monday morning , are you available

Need help with these two

I'm working I'm afraid. Monday mornings are always very busy and I won't be finished for several hours. Sorry :-(

@JohnRennie Thats ok ! Will catch you in night or tomorrow

@Jasmine for the mirror you just need find focal length, if it behave as concave or convex it depend upon it, s focal length sign.

1/f_r=2/f_1+2/f_2

And then rest differentiate the focal length.

@YuvrajSingh... I am not getting the correct answer

@JohnRennie Hello !

If you are free then can you please help with those

@Jasmine hi :-)

@JohnRennie Hello :-)

What's the problem with the irst question - Q90?

@Jasmine hello ?

@JohnRennie Saurav's statement, there will be some component of Impulse along x-axis as well

There is a horizontal frictional force.

@JohnRennie no friction

The diagram says "rough"

It says "neglect friction between the ball and wedge", but nothing about neglecting friction between the wedge and the floor.

@JohnRennie it also says neglect friction between ball and wedge

@JohnRennie Oh yes , didn't pay attention

So in that case Saurav is correct

So there is an external horizontal force acting since we are treating the floor as fixed.

Saurav is correct.

@JohnRennie answer given is (D)

Probably key mistake

I think C is correct.

What about this ↓

My approach was using equivalent focal length concept

But I didn't get the correct answer

@JohnRennie Okay may be wrong answer given

You need to work out the combined focal length of the two lenses, which is straightforward if a bit tedious.

@JohnRennie Yes I did and got it as 60

Then you need to calculate $v$, then calculate $dv/du$.

@JohnRennie yes , but I am not getting the right answer

Probably I am making a mistake

Then the speed of the image is $dv/dt = dv/du \times du/dt$

Is it possible to analyse if the resultant focal length is positive or negative without actually solving the question

Remember that the light passes through the (combined) lenses twice, so the total power is doubled ...

@JohnRennie Yes

Yes because the converging lens has a positive power and the diverging lens a negative power. When you add them you just have to see if the result is positive or negative.

It's going to be negative isn't it, because the lenses have the same radius and the diverging lens has a higher refractive index.

So the combined lens is diverging.

We can go through the calculation if you want ...

@JohnRennie yes

Probably I have made mistake in sign convention

@JohnRennie yes sure if it is ok with you

@Jasmine I said the same in few words, because process is simple.

I get the combined focal length to be -120 cm.

So because the light passes through the lens twice that gives a total focal length of -60 cm, which is the same as you got.

@JohnRennie I got +60

The lens is diverging isn't it?

Or am I getting the sign convention wrong?

@JohnRennie Plugging in that focal length gives the correct answer

The problem is that I learned the older sign convention that isn't used now, and I sometimes struggle with the current sign convention. Especially when mirrors are involved.

I would have just ignored the mirror. All the mirror does is reverse the direction of motion so the magnitude won't change, and doing the calculation for simple lens without a mirror is a lot easier.

@YuvrajSingh... can you please just calculate once and tell the equivalent focal length

@JohnRennie The sign convention I have learned is that distance measured opposite to direction of light is negative

@JohnRennie For the plano concave lens↓

I get 1.8i as the answer

(1.5-1)(0-(1/-20))

@JohnRennie yes correct

I am not going to do calculation but I can give sign convention details that make easy for you.

Is that OK?

@Jasmine

Will the radius for plano concave lens be taken as positive

@YuvrajSingh... Ok

So what you have written for first interface is correct.

Now for second interface medium you know.

@JohnRennie chat.stackexchange.com/transcript/message/53629342#53629342

Is it what you got as well

Take R1 as positive and R2=as minus infinity.

@YuvrajSingh... why R1 positive

Oops sorry R1 is negative and R2=+ infinity

@Jasmine I wouldn't worry too much about the sign conventions in the lensmakes equation, because the plano convex lens is obviously converging so it has a positive focal length (f = +60).

@JohnRennie correct sir!

Likewsie the plano concave lens is obviously diverging (f = -40 cm)

But sign convention are essential for jee problems.

@JohnRennie

@JohnRennie yes

Add them together and you get f = -120 cm

I dont know where am I getting the convention wrong

Then double the lens and you get f = -60 cm.

@JohnRennie Ok

@YuvrajSingh... that will give you +40

@Jasmine converging lenses are always positive and diverging always negative aren't they?

@JohnRennie I dont know , but my teacher advised me to always use the sign convention

@Jasmine have you see

@JohnRennie yes..

@YuvrajSingh... If you agreed to this , it is coming out to be -60

3/2-1(-1/20)

@JohnRennie the light is entering from the plane surface and not the curved surface in plano concave lens

@Jasmine are we agreed that R1 = infinity and |R2| = 20 for the planoconcave lens with the direction of light drawn?

@Jasmine I think you over doing or more focusing, thing is you need to see incoming rays give your sign convention according to that.

It still gonna diverge it , but I guess the focal length here is negative

@JohnRennie yes

@Jasmine and the centre of the concave surface is in the direction of travel of the light so the sign of R2 is positive. Yes?

If you think incoming rays are coming from positive x axis

Then define and radius of curvature as radius which inscribed the incoming ray.

@JohnRennie Why so

@Jasmine the rule is: a positive R indicates a surface's center of curvature is further along in the direction of the ray travel, while negative R means that rays reaching the surface have already passed the center of curvature.

@JohnRennie Well I wasnt aware of such rule

Now you are aware if it :-)

@JohnRennie I said same rule in my last. Comment copy that.

The only rule I have been taught is take the distance measured from centre of curvature in direction of light to be positive

@JohnRennie thats ok,

I don't but I was taught something different

Is it a misconception ^

To measure the radius of curvature we are going in direction opposite to light

So what do you use for the lensmaker's equation? The one I know is $$1/f = (n-1)(1/r_1 - 1/r_2) $$

@JohnRennie yes I used the same

The only thing I am doing wrong is taking r2 as negative

But doesn't that mean your focal length is going to come out positive for the planoconcave lens?

Which according to what I have been taught should be correct

@JohnRennie :-( for planoconcave I am getting 60

How does that come out as -60?

Or did you mean the planoconvex lens is -60?

I am messing up :-(

Just a typo :-)

@JohnRennie I am getting it as +60

@Jasmine for which lens?

@JohnRennie planoconcave

For planoconcave I am getting +40

For planoconvex I am getting -60

OK. I agree except for the signs. That's because the sign convention you are using seems wrong to me.

As far as I know a diverging lens always has a negative focal length and a converging lens always has a positive focal length.

@JohnRennie Sorry for troubling you, gonna study sign convention from HC Verma now

OK. I have to go but I'll be around tomorrow as usual.

But this was exactly what I was taught

@JohnRennie Ok see you tomorrow :-)

Oh got my mistake !!

No matter what we gonna measure we have to measure from optical centre !!