Alaser beam enters a 20.0 cm thick glass window at an angle of 41.0° from the normal. the index of refraction of the glass is 1.49. at what angle from the normal does the beam travel through the glass? how long does it take the beam to pass through the plate?

Angle of refraction = 26.12°, time taken = 0.89 ns

Explanation:

Using Snell's law as:

Where,  

 is the angle of incidence  ( 41.0° )

is the angle of refraction  ( ? )

is the refractive index of the refraction medium  (glass, n=1.49)

is the refractive index of the incidence medium (air, n=1)

Hence,

Angle of refraction= sin⁻¹ 0.4403 = 26.12°.

Also,

The distance it has to travel = 20.0 cm × cos 26.12° = 17.9575cm

Also,

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

Speed of light in vacuum = 3×10¹⁰ cm/s

Speed in the medium is:

v = c/n = 3×10¹⁰ cm/s / 1.49 = 2.0134×10¹⁰ cm/s

The time taken is:

t = d/s = 17.9575 cm / 2.0134×10¹⁰ cm/s = 8.92×10⁻¹⁰ s ≅ 0.89×10⁻⁹ s

Also,

1 ns = 10⁻⁹ s

So, time taken = 0.89 ns

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