Show that the deviation angle \(\delta\) for a ray striking a thin converging lens at a distance \(d\) from the principal axis is given by \(\delta=d l f .\) Therefore, a ray is bent through an angle \(\delta\) that is proportional to \(d\) and does not depend on the angle of the incident ray (as long as it is paraxial). [Hint: Look at the figure and use the small-angle approximation \(\sin \theta=\tan \theta \approx \theta(\text { in radians) } .]\)

Let's start by drawing a ray diagram for the given situation. The incident ray enters the lens at a distance \(d\) from the principal axis, forms an angle \(\alpha\) to the principal axis, and after passing through the lens, deviates by an angle \(\delta\) to form an angle \(\beta\) with the principal axis. We will also denote the focal length of the lens as \(f\).

For a ray passing through a thin converging lens, we can use the thin lens formula:$$ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} $$where \(u\) is the object distance, \(v\) is the image distance, and \(f\) is the focal length. Since the incident ray is paraxial (almost parallel to the principal axis), we can assume \(u \approx \infty\) as it is coming from far away. Then, the thin lens formula simplifies to:$$ \frac{1}{f}=\frac{1}{v} $$Solving for \(v\), we get:$$ v=f $$This means that the image is formed at a distance equal to the focal length of the lens.

Now that we have the image distance, we can find the angles involved. Since the object distance is very large (\(u \approx \infty\)), we can approximate the angle \(\alpha\) using the small-angle approximation:$$ \alpha =\frac{d}{u} \approx \frac{d}{\infty}=0 $$Then, we can approximate the angle \(\beta\) using the small-angle approximation:$$ \beta =\frac{d}{v} \approx \frac{d}{f} $$Since we have found \(\alpha\) and \(\beta\), we can now find the deviation angle \(\delta\):$$ \delta=\beta-\alpha\approx\frac{d}{f}-0=\frac{d}{f} $$Thus, the deviation angle for a ray striking a thin converging lens at a distance \(d\) from the principal axis is given by \(\delta=d/f\). As a result, a ray is bent through an angle \(\delta\) that is proportional to \(d\) and does not depend on the angle of the incident ray (as long as it is paraxial).