Question

A laser beam enters a 16.5 cm thick glass window at an angle of 58.0° from the normal. The index of refraction of the glass is 1.47. At what angle from the normal does the beam travel through the glass?

Answer 1

Answer:

= 35.23°

time taken = 0.9898 ns

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,  

 is the angle of incidence  ( 58.0° )

is the angle of refraction  ( ? )

is the refractive index of the refraction medium  (glass, n=1.47)

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

Hence,

$1\times {sin58.0^0}={1.47}\times{sin\theta_r}$

Angle of refraction= sin⁻¹ 0.5769 = 35.23°

The distance it has to travel

d =  = 16.5 cm / cos 35.23° = 20.20cm

Also,

n = c/v.

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

Speed in the medium is:

v = c/n = 3×10¹⁰ cm/s / 1.47

  = 2.0408 × 10¹⁰ cm/s

The time taken is:

t = d/s

  = 20.20 cm / 2.0408×10¹⁰ cm/s

  = 9.898 × 10⁻¹⁰ s

   = 0.9898 × 10⁻⁹ s

Also,

1 ns = 10⁻⁹ s

So, time taken = 0.9898 ns

Answer 2

Answer:

using Snells law

Oi = angle of incidence = 58.0°

    ni = index of refraction of air = 1.0003

    nr = index of refraction of glass = 1.47

    c = speed of light in vacuum = 3 x 10^8 m/s

    Or = angle of refraction = ?

ni(sinOi) = nR (sinOr)

ni( sinOi)/ nR = sinOr

arcsin(ni(sin0i))/nR = Or

arcsin( 1.0003(sin58.0)) / 1.47

Or = 35.25°

Explanation: