Test Index

WAEC Physics 2010 Paper

Question : 29

Total: 50

A ray of light travels from air to water. The refeactive index of water is given by the expression

$\frac{\text{velocity in water}}{\text{velocity in air}}$
$\frac{\text{sine of angle of refraction}}{\text{sine of angle of incidence}}$
$\frac{\text{apparent depth}}{\text{real depth }}$
$\frac{\text{wavelength in air}}{\text{wavelength in water}}$

Solution:

Concept:

Refractive index compares the speed of light in air with its speed in water, and this ratio also equals the wavelength ratio because frequency stays constant.

Explanation:

When light travels from air to water,

\text{refractive index of water} = \frac{\text{velocity of light in air}}{\text{velocity of light in water}}

Since

v = f\lambda

and frequency

f

remains unchanged during refraction,

\frac{\text{velocity in air}}{\text{velocity in water}} = \frac{\lambda_{\text{air}}}{\lambda_{\text{water}}}

Thus, the refractive index is expressed as the ratio of wavelength in air to wavelength in water.

Option D correctly represents this expression.

Answer:

\frac{\text{wavelength in air}}{\text{wavelength in water}}

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