https://isaacphysics.org/questions/changing_perspective?board=732fde2c-5cd8-49d7-9afc-86d8f3fb3ba6&stage=a_level

Im stuck on part c of this question

I don't really understand how to work out the distance from the optical axis as the initial rays never cross the axis - so I dont really know how to use the lens equation.

Any pointers would be greatly appreciated.

Im stuck on part c of this question

I don't really understand how to work out the distance from the optical axis as the initial rays never cross the axis - so I dont really know how to use the lens equation.

Any pointers would be greatly appreciated.

Working:

I initially thought that the distance would just be 2*5sin45 from looking at the diagram as I assumed the 2 triangles were the same.

I initially thought that the distance would just be 2*5sin45 from looking at the diagram as I assumed the 2 triangles were the same.

(edited 1 month ago)

Original post by mosaurlodon

Working:

I initially thought that the distance would just be 2*5sin45 from looking at the diagram as I assumed the 2 triangles were the same.

I initially thought that the distance would just be 2*5sin45 from looking at the diagram as I assumed the 2 triangles were the same.

The location of the virtual image (is the object for the converging lens) but the distance between converging lens and the virtual image has changed.

So the object’s distance from the converging lens is NOT 15*cos(45°), using this value, you are changing the location of the virtual image.

You should have worked out the location of the virtual image in part B.

The diagram provided by Isaac Physics has something missing that I believe is meant for the students to add.

The so-called new rays diagram for the problem looks like the diagram shown below but flipped upside down and the problem is asking for “height of A’B’ ” NOT d shown in your diagram.

Don't need similar triangles.

Thank you - I overlook adding the other length.

So in part b I worked out the image is formed 450cm to the right of the converging lens.

and to the left of the converging lens the distance to the virtual image is 45+5=50cm.

so u=50+5cos45

and angle alpha is the same in both triangles.

So you can work out alpha using trig and use lens equation to find v and find d?

So in part b I worked out the image is formed 450cm to the right of the converging lens.

and to the left of the converging lens the distance to the virtual image is 45+5=50cm.

so u=50+5cos45

and angle alpha is the same in both triangles.

So you can work out alpha using trig and use lens equation to find v and find d?

Original post by mosaurlodon

Thank you - I overlook adding the other length.

So in part b I worked out the image is formed 450cm to the right of the converging lens.

and to the left of the converging lens the distance to the virtual image is 45+5=50cm.

So in part b I worked out the image is formed 450cm to the right of the converging lens.

and to the left of the converging lens the distance to the virtual image is 45+5=50cm.

This is indeed for part B.

Original post by mosaurlodon

so u=50+5cos45

and angle alpha is the same in both triangles.

So you can work out alpha using trig and use lens equation to find v and find d?

so u=50+5cos45

and angle alpha is the same in both triangles.

So you can work out alpha using trig and use lens equation to find v and find d?

But the above is not for part C.

The “object’s distance” from the converging lens is NOT u = 50 + 5cos(45°)

Read the question Part C again. I interpret the question wrongly.

There is no need of alpha. See the diagram I have shown but flipped upside. You don’t need other angles, only 45° is needed and no other triangle other than the right-angled triangle with x = 5.

Read the part C carefully to work out u and then apply the lens equation to work out v….

Original post by mosaurlodon

Thank you - I overlook adding the other length.

So in part b I worked out the image is formed 450cm to the right of the converging lens.

and to the left of the converging lens the distance to the virtual image is 45+5=50cm.

so u=50+5cos45

and angle alpha is the same in both triangles.

So you can work out alpha using trig and use lens equation to find v and find d?

So in part b I worked out the image is formed 450cm to the right of the converging lens.

and to the left of the converging lens the distance to the virtual image is 45+5=50cm.

so u=50+5cos45

and angle alpha is the same in both triangles.

So you can work out alpha using trig and use lens equation to find v and find d?

If possible, please upload the diagram in jpg form, it is rather difficult to reply with png diagram. On my side, your png diagram looks like this (I just show 2 of them below):

...

It is not easy to write reply.

Ok, thank you for both of your replies and sorry about the image - I will keep that in mind.

I've managed to get the answer as 45cm but I dont know if I did it with the right method?

I used this image: https://isaacphysics.org/images/content/concepts/physics/figures/concepts_lenses_lens_ray_diagram.svg

I said u = 45+5cos45 (not u=50+5cos45 - misunderstood by me sorry about that)

h = 5sin45

worked out v using 1/f - 1/u and got v = 620cm

and used equation h/u = x/v (similar triangles)

and got x as 45cm.

is this correct?

I've managed to get the answer as 45cm but I dont know if I did it with the right method?

I used this image: https://isaacphysics.org/images/content/concepts/physics/figures/concepts_lenses_lens_ray_diagram.svg

I said u = 45+5cos45 (not u=50+5cos45 - misunderstood by me sorry about that)

h = 5sin45

worked out v using 1/f - 1/u and got v = 620cm

and used equation h/u = x/v (similar triangles)

and got x as 45cm.

is this correct?

Original post by mosaurlodon

Ok, thank you for both of your replies and sorry about the image - I will keep that in mind.

I've managed to get the answer as 45cm but I dont know if I did it with the right method?

I used this image: https://isaacphysics.org/images/content/concepts/physics/figures/concepts_lenses_lens_ray_diagram.svg

I said u = 45+5cos45 (not u=50+5cos45 - misunderstood by me sorry about that)

h = 5sin45

worked out v using 1/f - 1/u and got v = 620cm

and used equation h/u = x/v (similar triangles)

and got x as 45cm.

is this correct?

I've managed to get the answer as 45cm but I dont know if I did it with the right method?

I used this image: https://isaacphysics.org/images/content/concepts/physics/figures/concepts_lenses_lens_ray_diagram.svg

I said u = 45+5cos45 (not u=50+5cos45 - misunderstood by me sorry about that)

h = 5sin45

worked out v using 1/f - 1/u and got v = 620cm

and used equation h/u = x/v (similar triangles)

and got x as 45cm.

is this correct?

Should be good.

Thank you! 😀

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