Question

1. Light of wavelength 620nm in air is incident on a surface of a flat piece of glass with an index of refraction of

1.50 with an angle of incidence of 30°.

a.What is the angle of refraction?

-Answer: 19.5°

b. Find the wavelength and frequency of light in glass.

Answer: 413nm, 4.84x10¹⁴Hz

c.Explain under what circumstances “total internal reflection” can be observed at the boundary between air

and this piece of glass. Be as specific as possible.

Answer 1

Angle of incidence i = 30 deg

Angle of refraction = r= ?

ref index = Sin(i)/Sin (r)

1.5 = sin 30/ sin(r)

sin(r) = sin 30/1.5 = 0.33

r = 19.5 degrees

b. wavelegth in air = 620nm

Velocity of light in air = 3X 10^8 m/s

Frequency = 3 X 10^8m/s / 620 x 10^(-9) mrs = 4.84 X 10^14 Hz

frequency remains at 4.84 X 10^14 Hz inside glass too.

However, velocity of light reduces inside the glass

refractive index = 1.5 = velocity of light in Air/vel of light in glass

Velocity of light in glass = 3 X 10^8 m/s/ 1.5 = 2 X 10^8 m/s

Wavelength in Glass = 2 X 10^8 m/s / 4.84 X 10^14 hz = 413 X 10^(-9) m = 413 nm

c. Total internal is a phenomenon which takes place when light transits from a denser medium to a lighter medium. In this case thus this phenomenon takes place when light travels from Glass to air.

Let us first start with low incident angles i and and progressively increase the value of i

Snells law governs the Refraction of light emerging from glass to air.

  

As we increase ,   also increases and a stage will come where   = 90 deg and the corresponding value of is called critical angle C

Thus C is that incident angle at which the refracted ray grazes along the surface of glass slab and hence if

i > C , there will be total internal reflection at the interface of Glass and Air.

In this case,

1.5 = 1/Sin C

Sin (C) = 1/1.5 = 41.8 degress

So for all i such that

i > 41.8 deg , we get Total Internal Reflection