Question

(a) The index of refraction for violet light in silica flint glass is 1.66, and that for red light is 1.62. What is the angular spread (in degrees) of visible light passing through a prism of apex angle 60.00 if the angle of incidence is 52.0°? See figure below. Deviation of - red light Visible light Angular spread Screen 4.3 4.3 vo (b) What If? What is the angular spread (in degrees) of visible light passing through a prism of apex angle 60.0° if the angle of incidence is 90°?

Answer 1

a)

If the angle of incidence is

52

For violet:  

1=1 n2 = 1.66

angle of refraction at first surface is $r_1$

Snell's law at the first surface,

1 Sin n2 r1 sin пi sin i n2 sin 52 1.66 - 28.3402 sin-1 sin r

The ray incident at an angle $r_2$ at the second surface. The two normals meet an angle

Ti + 12 + 120º = 180°

$r_2=60\degree-r_1=31.6598\degree$

Snell's law for the second surface,

${\frac{sin r_2}{\sin e_v}=\frac{n_1}{n_2}\Rightarrow e_v=\sin^{-1}\left ( \frac{n_2}{n_1}\sin r_2 \right )=\sin^{-1}\left (1.66\sin 31.6598\degree \right )=60.6092\degree}$

Exit angle of violet is $e_v=60.6092\degree$

For red: ray

$n_2=1.62$

angle of refraction at first surface is $r_1$

Snell's law at the first surface, $\frac{\sin i}{\sin r_1}=\frac{n_2}{n_1}\Rightarrow r_1=\sin^{-1}\left ( \frac{n_1}{n_2}\sin i \right )=\sin^{-1}\left ( \frac{1}{1.62}\sin 52\degree \right )=29.1060\degree$

The ray incident at an angle $r_2$ at the second surface. The two normals meet an angle

Ti + 12 + 120º = 180°

$r_2=60\degree-r_1=30.8940\degree$

Snell's law for the second surface,

${\frac{sin r_2}{\sin e_r}=\frac{n_1}{n_2}\Rightarrow e_r=\sin^{-1}\left ( \frac{n_2}{n_1}\sin r_2 \right )=\sin^{-1}\left (1.62\sin 30.8940\degree \right )=56.2832\degree}$

Exit angle of red is $e_r=56.2832\degree$

Angular spread is $e_v-e_r=60.5092\degree-56.2832\degree}=4.33\degree$

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b)

If the angle of incidence is

206 = ?

For violet:  

1=1 n2 = 1.66

angle of refraction at first surface is $r_1$

Snell's law at the first surface,

1 sin - sin 90° 1.66 = 37.0427° r1 = si1 =sin-1 sin ri n2

The ray incident at an angle $r_2$ at the second surface. The two normals meet an angle

Ti + 12 + 120º = 180°

r2 = 60° - 11 = 22.9573

Snell's law for the second surface,

nr2 = "1= €, = sin-1 (1-2 sin r2) = sin-2 (1.66 sin 22.9573') = 40.35148° sine, n2 ni

Exit angle of violet is

ey = 40.35148

For red:

n1= 1 , n2 = 1.62

If the angle of incidence is
206 = ?
, angle of refraction at first surface is $r_1$

Snell's law at the first surface,

sin sin 90° 1.62 = 38.11806 r1 = si1 =sin-1 sin ri n2

The ray incident at an angle $r_2$ at the second surface. The two normals meet an angle $120\degree$

Ti + 12 + 120º = 180°

T2= 60 r1 = 21.88194°

Snell's law for the second surface,

= 37.14012 п2 sin r2 sin r2 sin(1.62 sin 21.881940) er=sin-1 п2 = пi sin er

Exit angle of violet is

e 37.14012

Angular spread is

40.35148 37.14012° = 3.211 E-e

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