& critical angle t
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Transcribed Image Text:(4.
Find the critical angle for lotal Internal
on the surface
Reflection of the light incident
between the air and the diamond at point P
a.) sindc =
sin-i
иг
ni
(sind) = 1.00
2.419) sint
0₁ = 24.4⁰°
nz
n₁
point
=
1.333
2.419
n, sind, = n₂ sin da
35° 0
b.) If the diamond is totally immersed IN H₂0, find the
Critical angle at the Diamond - Water interface.
sin' (sinds)=
(
sin
33°
(2.419) sin(+6)= (1.33) sin O₂
sint (sin od=
air
12.419 (sin (1160)
1.33
(1²60)) s.nl
airnz 1.000
Diamond (n) = 2.6419
water = 1.333
·Diamond
A
Oc= 33.4°
C.) assuming the light ray remains perpendiular on entering the
diamond, at what angle would the Light begin to emerge
at point & if the diamond is rotated about point
(Ⓒ of retraction) use Snell! diamond in the water (33.39⁰)
25°-33.4: 1.6°
Todran > 33.4
Ог
if & were a bit less
1
than 33.4 Like 33.3
-187.07°
2
the the Or in water
w
would be 85.07
or even 33.39
would give Or
of 87.07
+
0,= $ 12,9° / +
2
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