# Modified plano-convex lens doubt

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1. Dec 16, 2014

I was thinking about the situation given my text about a plano convex lens which was produced with a manufacturing defect. It's plane surface is tilted outwards by a small angle 'z'. In the text its written that when a parallel light beam enters the lens parallel to x axis , it will still be able to focus it at:
$(R/(u-1) , -zR)$ where u is refractive index of lens and R is the radius of curvature of curved surface of lens and lens is at origin with rays along x axis starting from negative infinity. How does this happen?

My analysis: I think the tilted surface is just for bending the ray(refraction) by a small angle 'r' and now we have Plano convex lens with parallel beam of light at an angle 'r'.

2. Dec 16, 2014

### Staff: Mentor

The plane surface of the lens is still flat, right? If so, what is -zR?

3. Dec 16, 2014

No. Plane surface is tilted at an angle z.

4. Dec 16, 2014

### Andy Resnick

Think about it this way- your lens is an ideal lens plus a wedge prism. does that help?

5. Dec 16, 2014

### Staff: Mentor

It's tilted, but it's not curved, right?

6. Dec 16, 2014

Yes. Its not curved.

7. Dec 16, 2014

Igot the x-coordinate part. But how do you get the zR?

8. Dec 16, 2014

### Staff: Mentor

Hmmm, I think your original analysis is pretty much correct. The tilted plane surface should just change the angle of the converging beam. I'm guessing -zR is the distance the new focal point is from the original focal point, but I'm not sure.

9. Dec 17, 2014

### Andy Resnick

How about using the small angle approximation sin z = z?