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The lens maker's formula takes into account the thickness of a lens. However, if one surface is planar, its focal length should be infinite. But that means that with the below formula, the thickness of a lens doesn't affect its focal length if one side is planar which isn't true.

What's wrong here? (assuming $n$ of air $= 1$)

$$\frac{1}{f} = (n - 1)\bigg(\frac{1}{r_1} - \frac{1}{r_2} + \frac{(n-1)d}{nr_1r_2}\bigg)$$

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  • $\begingroup$ The focal length of plane surface is infinite not of the lens because one surface still has curvature. $\endgroup$
    – SarGe
    Commented May 12, 2020 at 10:38

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If one surface of a thick lens is planar, then the effective focal length does not change with thickness. However, the back focal distance does change, and you may be confusing this with effective focal length. The back focal distance is the distance from the last surface of the lens to the image plane.

For example, if the front surface of the lens is planar, then the back surface is the only surface that has optical power. In that case, the focal length and the back focal distance are the same.

If the back surface is planar, then only the front surface has power. The focal length does not change with lens thickness, but the back focal distance decreases if the lens thickness increases. Having a thicker lens in this case is just the same as inserting an extra piece of plane-parallel piece glass between the lens and the image plane.

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  • $\begingroup$ so if you were to calculate the lateral shift, would that be the difference in the back focal length and effective focal length? $\endgroup$
    – ganuwoahh
    Commented May 13, 2020 at 9:30
  • $\begingroup$ If by lateral shift, you mean the height of an off-axis point at the image plane, then no -The height of an off-axis field point at the image plane is determined by the relation h=f*tan(theta), where h is the height, f is the effective focal length, and theta is the field angle. (assuming an object at infinity for simplicity). Putting a block of glass after the lens will change where the image plane is, but the off-axis rays still image to the same off-axis position at the image plane. $\endgroup$
    – JB2
    Commented May 13, 2020 at 11:45
  • $\begingroup$ i mean like lateral displacement due to a glass slab. also please dumb down your explanations im only a high school student :( $\endgroup$
    – ganuwoahh
    Commented May 13, 2020 at 13:15

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