Reading an Engraving Through a Perspex Window
An inspector must read a serial number engraved on the inner face of a 12 mm thick cast Perspex window (refractive index 1.49) and examine a 0.15 mm scratch on the same face. Her eye's near point is 250 mm. She has a plano-convex magnifier of focal length 60 mm and a loupe of focal length 25 mm. The window is edge-lit: a lamp is pressed against the edge so light is trapped inside the Perspex and scattered out only where a scratch or an engraving disturbs the surface. The window has just been cleaned with water and is still wet. She plans to use the 60 mm magnifier for both tasks and to judge the engraving depth from how far she has to move the lens.
1. Where does the engraving appear to be, and what does that mean for her depth judgement?
Read the worked answer
Apparent depth = real depth / n = 12 / 1.49 = 8.1 mm.
The engraving therefore appears about 4 mm closer to the observer than it really is, i.e. just under the outer surface. The apparent scale factor is 1/1.49 = 0.67.
Any lens-position or focusing measurement made from the outer face measures the apparent depth, not the true one, so a 'shallow engraving' conclusion drawn that way is an artefact of refraction. To recover the true depth, multiply the apparent distance by n: 8.1 × 1.49 = 12 mm. If the real question is 'is the engraving deep enough?', a lens position is the wrong instrument - use a dial indicator referenced to the inner face or a shadowgraph.
2. Compare the two lenses for the 0.15 mm scratch. Give the magnifications and the trade-off.
Read the worked answer
Angular magnification with the image at the near point (D = 250 mm): M = 1 + D/f.
- Magnifier, f = 60 mm: M = 1 + 250/60 = 5.2×
- Loupe, f = 25 mm: M = 1 + 250/25 = 11×
With the eye relaxed and the object at the focal point, M = D/f: 4.2× and 10× respectively - smaller magnification but no eye strain.
Resolution: the unaided eye resolves roughly 0.1 mm at 250 mm, so a 0.15 mm scratch is marginal without help; at 11× it is comfortably resolved, at 5.2× it is not reliably so.
Trade-off: higher magnification means a smaller field of view and a shorter working distance. The loupe must be held close to the surface, which is a real risk on a wet window (contact, contamination, and a water film on the lens). Practical answer: 60 mm lens to find the serial number, 25 mm loupe - after drying - to grade the scratch.
3. Explain why the wet window spoils the edge-lit inspection, using the critical angle.
Read the worked answer
Perspex in air: sin C = n₂/n₁ = 1/1.49 = 0.671 → C = 42.2°.
Light travelling inside the window and striking a dry Perspex–air face at more than 42.2° from the normal is totally internally reflected and stays trapped. It escapes only where a scratch or engraving scatters it, so the defect appears bright on a dark background - exactly the contrast the technique depends on.
With a water film (n = 1.33) on the outer face: sin C = 1.33/1.49 = 0.893 → C = 63.2°.
A far larger range of ray angles now refracts out at the wet face, so the whole surface glows and the scratch disappears into the background. The window must be dried before the edge-lit examination - the technique is defeated by the coupling medium, not by the inspector's eyesight.