Question 29: Focus of a Defective (Tilted) Lens
1. Analyzing the Defect
The lens is a plano-convex lens with radius of curvature $R$ and refractive index $\mu$. Due to a manufacturing defect, the plane surface is tilted by a small angle $\alpha$.
This tilt effectively turns the lens into a combination of a standard plano-convex lens and a thin prism of angle $\alpha$.
2. Calculating the Deviation
A thin prism with angle $\alpha$ and refractive index $\mu$ produces a deviation $\delta$ given by:
$$ \delta = (\mu – 1)\alpha $$Since the plane surface is tilted, incident parallel rays striking the flat surface undergo refraction. The wavefront emerging from the flat surface is tilted by the angle $\delta$ relative to the vertical. Effectively, the rays enter the spherical part of the lens at an angle $\delta$ with the principal axis.
3. Coordinates of the Focus
X-Coordinate (Focal Length):
The focal length of a plano-convex lens is determined by the Lens Maker’s Formula:
$$ \frac{1}{f} = (\mu – 1)\left( \frac{1}{R} – \frac{1}{\infty} \right) \implies f = \frac{R}{\mu – 1} $$
So, the x-coordinate is $x = \frac{R}{\mu – 1}$.
Y-Coordinate (Lateral Shift):
A parallel beam incident at an angle $\delta$ focuses on the focal plane at a height $y$ given by:
$$ y = -f \tan \delta \approx -f \delta $$
(The negative sign indicates the shift is in the direction opposite to the tilt normal or consistent with the prism deviation direction).
Substituting $f$ and $\delta$:
$$ y = -\left( \frac{R}{\mu – 1} \right) \times (\mu – 1)\alpha $$
$$ y = -R \alpha $$
Final Answer
The defective lens will focus the beam at coordinates:
$$ \left( \frac{R}{\mu – 1}, \; -R\alpha \right) $$