A prism having retracting angle 5 degree has refractive index 1.5 the
a
ngle of minimum deviation is -
Answers
Answered by
0
Answer:
Let δ
m
be the angle of minimum deviation A=δ
m
A=δ
m
μ=
sin(
2
A
)
sin
2
(A+δ
m
)
1.5=
sin(
2
A
)
sin(
2
A+A
)
1.5=
sin(
2
A
)
sin2(
2
A
)
=
sin(
2
A
)
2sin
2
A
cos
2
A
1.5=2cos
2
A
0.75=cos
2
A
2
A
=cos
−1
(0.75)
2
A
=41
o
A=82
o
Answered by
0
Answer:
A=82
Explanation:
Let δ
m
be the angle of minimum deviation A=δ
m
A=δ
m
μ=
sin(
2
A
)
sin
2
(A+δ
m
)
1.5=
sin(
2
A
)
sin(
2
A+A
)
1.5=
sin(
2
A
)
sin2(
2
A
)
=
sin(
2
A
)
2sin
2
A
cos
2
A
1.5=2cos
2
A
0.75=cos
2
A
2
A
=cos
−1
(0.75)
2
A
=41
o
A=82
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