# Modified plano-convex lens doubt

In summary, 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).

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'.

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

Drakkith said:
The plane surface of the lens is still flat, right? If so, what is -zR?
No. Plane surface is tilted at an angle z.

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

No. Plane surface is tilted at an angle z.

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

Drakkith said:
It's tilted, but it's not curved, right?
Yes. Its not curved.

Andy Resnick said:
Think about it this way- your lens is an ideal lens plus a wedge prism. does that help?
Igot the x-coordinate part. But how do you get the zR?

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.