Calculate Separation of Two Biconvex Lenses

[nbb]

I am trying past paper exam questions for my wave optics paper and have got stuck on this question.

Two identical biconvex lenses of refractive index 1.5 and with radius of curvature of 20cm on each side are to be used to produce a real erect image of incoming parallel rays.
If the image is to be formed 30cm behind the second lens, calculate the separation of the lenses.

So far, I've calculated the focal length of the lenses (0.6m i think), and I think the separation must be less than this, but I'm not sure where to go next.

Any suggestions welcome.

Thanks.

 

[maverick280857]

Parallel incidence on either lens means the refracted rays will pass through the focus of the lens. Assume that two lens are separated by x units. You know the focal length of the first length (f). Can you take it from here?

 

[nbb]

Thanks for the prompt reply.

I've used 1/f = (n-1)(1/R1 - 1/R2) with R2 as a negative value to find f to be 0.2m. Which is not what I thought before.

Then I've used 1/u + 1/v = 1/f for the second lens with v = 0.3m to calculate u for the second lenses to be 0.6m.

If the image after the second lens is to be erect(ie above the optical axis) then the 'image' (is that what it's called?) between the two lenses must be below the optical axis at a distance of 0.6m from the second lens, and twice the size of the final image.

If parallel beams, which are erect (above optic axis) come from before the first lens and then are refracted through the focal point and then continue below the optic axis until the point when they are 0.6m from the second lens, this then means( i think) that the distance x between the lenses is 0.6m + 0.2m + the horizontal distance it travels between the focal point of the first lens and the point when it is 0.6m from the second lens.

If this is the right, then is the next step to use similar triangles to find this distance?

 

[maverick280857]

Thanks for the prompt reply.

I've used 1/f = (n-1)(1/R1 - 1/R2) with R2 as a negative value to find f to be 0.2m. Which is not what I thought before.

Then I've used 1/u + 1/v = 1/f for the second lens with v = 0.3m to calculate u for the second lenses to be 0.6m.

Good so far...

If the image after the second lens is to be erect(ie above the optical axis) then the 'image' (is that what it's called?) between the two lenses must be below the optical axis at a distance of 0.6m from the second lens, and twice the size of the final image.

Well I've not worked out your question numerically but you're probably right. The parallel rays intersect the principal axis of the first lens before the second lens only if the distance between the two lens is greater than the focal length of the first lens.

If parallel beams, which are erect (above optic axis) come from before the first lens and then are refracted through the focal point and then continue below the optic axis until the point when they are 0.6m from the second lens,

There you have it

this then means( i think) that the distance x between the lenses is 0.6m + 0.2m + the horizontal distance it travels between the focal point of the first lens and the point when it is 0.6m from the second lens.

If this is the right, then is the next step to use similar triangles to find this distance?

Since you do not know what the distance between the two lens is, you can take it as x and then consider the object for the second lens to be located at (x-f) units from the second lens (it could be a negative number in which case this is a virtual object). [You're not using the cartesian convention right?]

The imaging equation for the second lens can be solved to get x. The second and first focal point terminology is good but its avoidable for such questions and I personally think you should avoid it till you are adept with simpler interpretations...a diagram for this problem might help for example.

Cheers
Vivek

 

[Andrew Mason]

If parallel beams, which are erect (above optic axis) come from before the first lens and then are refracted through the focal point and then continue below the optic axis until the point when they are 0.6m from the second lens, this then means( i think) that the distance x between the lenses is 0.6m + 0.2m + the horizontal distance it travels between the focal point of the first lens and the point when it is 0.6m from the second lens.

If this is the right, then is the next step to use similar triangles to find this distance?

The easiest way to look at this problem is to think of the second lens as focusing an image that originates at the focal point of the first lens. Just use the lens equation to determine the object distance (distance from focal point of the first to the second lens):

[tex]\frac{1}{o} + \frac{1}{i} = \frac{1}{f}[/tex]

[tex]\frac{1}{o} = 1/.2 - 1/.3 = .1/.06 [/tex]

so [itex]o = .6[/itex]

Add that to the focal length to find the total separation = .8 m.

AM

 

[nbb]

Thanks, that's great.