vi) Critical angle between glass and air becomes (
ag = 3/2):
a) 41 48 b) 45 40 c) 35 20 d) 30 40
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Correct option is
A
62
∘
28
′
Critical angle sinθ
c
=
n
1
We get n=
sinθ
c
1
We will use the relation 1
o
=60
′
For glass :
Critical angle θ
c
=41
o
48
′
=41
o
+
60
48
=41
o
+0.8
o
=41.8
o
Refractive index for glass n
g
=
sin(41.8)
1
=
0.66
1
=1.5
For water :
Critical angle θ
c
=48
o
36
′
=48
o
+
60
36
=48
o
+0.6
o
=48.6
o
Refractive index for water n
w
=
sin(48.6)
1
=
0.75
1
=1.33
For water-glass interface :
Critical angle sinθ
c
=
n
g
n
w
=
1.5
1.33
=0.887
⟹ θ
c
=sin
−1
(0.887)=62.46
o
We get θ
c
=62
o
+ (0.46×60)
′
=62
o
28
′
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