A beam of laser light with wavelength 612 nm is directed through a slab of glass

Question

A beam of laser light with wavelength 612 nm is directed through a slab of glass having index of refraction 1.5.
(a) For what minimum incident angle would a ray of light undergo total internal reflection?
1Your answer is incorrect.°

(b) If a layer of water is placed over the glass, what is the minimum angle of incidence on the glass-water interface that will result in total internal reflection at the water-air interface? (The beam of light is directed upward through the slab of glass, through the layer of water, and experiences total internal reflection at the water-air interface on top.)
2°

Explanation / Answer

The total internal reflection will occur when light beam is passed through denser medium to rarer medium for the incident angle which results a refracted angle of 90 1) The refractive index is n = 1.5 Applying Snell's law at the air glass interface, nsini = sin90 (1.5)sini = 1 i = 41.81o 2) The refractive index of the water is n2 = 1.33 Applying Snell's law at the air water interface, n2sini' = sin90 (1.33)sini' = 1 i' = 48.75o The ray from the glass is refracted at r = i' = 48.75o Applying Snell's law at the glass water interface, n1sini = n2sin48.75 (1.5)sini = (1.33)sin48.75 i = 41.81o 2) The refractive index of the water is n2 = 1.33 Applying Snell's law at the air water interface, n2sini' = sin90 (1.33)sini' = 1 i' = 48.75o The ray from the glass is refracted at r = i' = 48.75o Applying Snell's law at the glass water interface, n1sini = n2sin48.75 (1.5)sini = (1.33)sin48.75 i = 41.81o Applying Snell's law at the glass water interface, n1sini = n2sin48.75 (1.5)sini = (1.33)sin48.75 i = 41.81o Applying Snell's law at the glass water interface, n1sini = n2sin48.75 (1.5)sini = (1.33)sin48.75 i = 41.81o

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