Help with A-Level physics Ray Diagrams

[olive iron]

It's simple I'm just slow

[heavy stratus]

1. Do not ping the Moderators, unless someone is breaking the rules.
2. Do not ping the Helper Moderators, unless there is a conflict between helpers.
3. Do not ping other members randomly for help.
4. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
5. Wait patiently for a helper to come along.
6. If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:

+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[olive iron]

I just have no clue what to do

[sharp belfry]

use the magnification formula to find a relation between v and u

[olive iron]

It's part ii

Not iii

Mb

[sharp belfry]

i mean, we have to get started somewhere

so yeah

height of image/ height of object = negative v/u

[olive iron]

Yes

So I need to find the image size

[sharp belfry]

no use the graph

take one box as 1 unit

[olive iron]

Or cm

[sharp belfry]

we need the ratio so units are just there, we wont be needing units

[olive iron]

Object height 3cm

[sharp belfry]

yeah

and image height is 1.5

[olive iron]

How

Do u know it's

/2

[sharp belfry]

yeah cuz there are equal number of smaller boxes above and below the point

so that makes the magnification -1/2

which means

v/u=1/2

now, we can make use of newtons formula

,w newtons formula for optics

[olive iron]

Okay

[olive iron]

Ion know this

[sharp belfry]

yeah im sending the derivation

one sec

nvm my bad

[olive iron]

It's alr

[sharp belfry]

so we can assume a point on the axis to be the pole of our lens

[olive iron]

Ye

[sharp belfry]

and that point has to be such that u=2v

[olive iron]

How come

[sharp belfry]

from the magnification relation we just derived

[olive iron]

Alr

[sharp belfry]

here

so we are gonna assume a coordinate system with foot of the object as our origin

and this time around, we are going to assume one big square box as 1 unit

alr?

[olive iron]

Ye

[sharp belfry]

which brings the foot of our image to be (5,0)

what we need is a point that divides 0,0 and 5,0 in the ratio 2:1

and for that we use the legendary section formula

which gives us 10/3,0 to be the foot of our lens

got it?

[olive iron]

Yes

[sharp belfry]

alr cool

so now we get the object distance to be negaive 10/3

and the image distance to be positive 5/3

[olive iron]

One thing

This question says use construction rays

[sharp belfry]

yeah we can do that once we get out numbers sorted out

[olive iron]

Alr

[sharp belfry]

now by substituting our values into lens formula

,w lens formula

huhh

,w lensmakers formula

so its 1/v+1/u = 1/f

where v = 5/3

u= -10/3

wait

i did smth wrong

[digital dagger]

@olive iron

:HelpIcon:| Help Reminder

Hello harold1402_, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.

[olive iron]

I forgor

I give up

+close

[digital dagger]

Please thank your Helpers before closing!

Please thank the helpers who assisted you by clicking the buttons below. You can thank each helper only once. Once you're done, click "Close Post" to close this thread.

[digital dagger]

Thank you for your feedback! Dr. Bored has been awarded 1 . They now have 1 . They have 3 daily left for today.