A parallel beam of light is incident on a face of 60° prism. By rotating the prism, the minimum angle of deviation is measured to be 40°. Find the refractive index of the prism material. If the prism is placed in water (refractive index=1.33), then what would be the new minimum angle of deviation?
Answers
Answered by
1
Answer:
Step by step explanation ____
Explanation:
The angle of the prism, 60
o
.
Minimum deviation δ=40
o
μ
′
is the refractive index of the material of the prism.
μ
′
=
sin(
2
A
)
sin(
2
A+δm
)
=1.5321
Let δ
1
be new minimum deviation, now it is put under water μ=1.33
⇒
μ
- μ
′
=
sin(
2
A
)
sin(
2
A+δm
)
⇒δ
1
=10.32
o
Answered by
0
Answer:
tomorrow will be help you okay dear
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