Question

A beam of light makes an angle of ?i = 18.0? with the normal of a slab of transparent material of t=1.10 mm-thickness. The index of refraction of the glass is n2 = 1.28.

A beam of light makes an angle of ?i = 18.0? with the normal of a slab of transparent material of t = 1.10 mm - thickness. The index of refraction of the glass is n2 = 1.28. a) What angle would the beam that leaves the glass surface make with the normal to the glass surface? Provide the answer in degrees. b) How far vertically (d) is the point at which the beam leaves the glass surface from the point at which the beam entered the glass? Provide the answer in mm.

a) What angle would the beam that leaves the glass surface make with the normal to the glass surface? Provide the answer in degrees.

b) How far vertically (d) is the point at which the beam leaves the glass surface from the point at which the beam entered the glass? Provide the answer in mm.

Answer 1

Answer 2

Answer 3

(a) The angle would the beam that leaves the glass surface make with the normal to the glass surface can be determined as follows:

Apply Snell's law:

$n_1sin\theta_i=n_2sin\theta_r$

Thus, the refracted angle is,

$\theta_r=sin^{-1}(\frac{n_1}{n_2}sin\theta_i)$

$\theta_r=sin^{-1}(\frac{1}{1.28}sin18) = 13.97^{\circ}$

Now this refracted ray incident on next surface. Thus, the refracted angle is,

$\theta_2=sin^{-1}(\frac{1.28}{1}sin13.97) = 18^{\circ}$

Hence, the ray leaves the glass surface with angle 18 degress.

(b)

Use trigonometry to find the verical height.

$tan\theta =\frac{d}{t}$

Thus, vertically distance is,