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Difficulty: MediumPhysics Fundamentals and Everyday Applications

A coin placed at the bottom of a beaker filled with water appears to be raised above its actual position when viewed from the air directly above. Which of the following optical phenomena is primarily responsible for this observed shift in apparent position?

  1. Refraction of light as it travels from an optically denser medium (water) into an optically rarer medium (air)Answer
  2. B
    Total internal reflection of light occurring at the water-air interface
  3. C
    Diffraction of light waves around the circular outer rim of the coin
  4. D
    Dispersion of light into its component wavelengths at the water surface

Answer

The apparent elevation of a submerged object in water is caused by the refraction of light as it passes from water into air.
Light rays reflected from the submerged object bend away from the normal when exiting water into air. Because human visual perception assumes light travels in unbent straight lines, tracing these refracted rays backward projects a virtual image higher than the object's actual physical location.

Step-by-Step Solution

1
Trace light rays from the object inside the water toward the surface.
Light rays travel from water (optically denser medium, refractive index nwater≈1.33n_{\text{water}} \approx 1.33) toward air (optically rarer medium, nair≈1.00n_{\text{air}} \approx 1.00).
For an object submerged in water to be visible to an external observer, light must cross the interface between the two media.
2
Apply Snell's law of refraction at the boundary.
The light rays bend away from the normal upon entering the air.
When light passes from a medium of higher refractive index to one of lower refractive index, the angle of refraction is greater than the angle of incidence.
3
Determine the position of the virtual image formed by the refracted rays.
Extrapolating the refracted rays backward in a straight line locates a virtual image above the object's actual depth.
The relation for apparent depth is given by d′=dnwaterd' = \frac{d}{n_{\text{water}}}, where dd is the real depth and d′d' is the shallower apparent depth.

Key Concept

Apparent Depth and Refraction at Plane Interfaces
Estimated Time:1m 0s
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Topics

Physics Fundamentals and Everyday Applications

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