A ray of light strikes a flat, 2.00-cm-thick block of glass ( n = 1.93) at an an
ID: 1836545 • Letter: A
Question
A ray of light strikes a flat, 2.00-cm-thick block of glass (n = 1.93) at an angle of = 32.6° with respect to the normal (see figure below).
(a) Find the angle of refraction at the top surface and the angle of incidence at the bottom surface.
°
(b) Find the refracted angle at the bottom surface.
°
(c) Find the lateral distance d by which the light beam is shifted.
cm
(d) Calculate the speed of light in the glass.
m/s
(e) Calculate the time required for the light to pass through the glass block.
s
Explanation / Answer
Thickness of the glass t = 2 cm
Index of refraction of glass n = 1.93
Angle of incidence i = 32.6 o
(a)The angle of refraction at the top surface and the angle of incidence at the bottom surface r = ?
From Snell's law , sin i / sin r = n
sin r = sin i / n
= sin 32.6 o / 1.93
= 0.2791
r = sin -1 (0.2791)
= 16.2 o
(b) the refracted angle at the bottom surface r ' = ?
From Snell's law sin r / sin r ' = 1 / n
sin r ' = n sin r
= 1.93 sin 16.2
= 0.5387
r ' = 32.6 o
(c) the lateral distance d by which the light beam is shifted d = ?
From figure , cos r = t / length of the bule ray in glass
Length of the blue ray in glass = t / cos r
= 2 cm / cos 16.2
= 2 cm / 0.96
= 2.082 cm
From figure, tan (i-r) = d / Length of blue ray in glass
d = Length of the blue ray in glass x tan ( i- r )
= 2.082 x tan (32.6 -16.2)
= 2.082 x tan(16.4)
= 0.6129 cm
(d) the speed of light in the glass , v = c / n
=(3 x10 8 ) /1.93
= 1.554 x10 8 m/s
(e) the time required for the light to pass through the glass block t ' = 2.082 cm / v
t ' =(2.082 x10 -2 m) / (1.554 x10 8 m/s)
= 1.339 x10 -10 s
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