A light ray travelling in medium 1 having refractive index 1 is
incident on a smooth flat slab of medium 2 having refractive
index 2 such that 2 > 1 . If the incoming ray makes an angle
1 with respect to normal, what is angle of refraction in the
medium 2, critical angle, NA, and acceptance angle?
Answers
Explanation:
When the ray travels from top to bottom of the slab,the refractive index of slab decreases (μ(x)=a−bx) and the ray will refract at 90° at critical angle.
When the ray travels from top to bottom of the slab,the refractive index of slab decreases (μ(x)=a−bx) and the ray will refract at 90° at critical angle.By Snell's law, 2×sin37°=μ(x)×sin90°
When the ray travels from top to bottom of the slab,the refractive index of slab decreases (μ(x)=a−bx) and the ray will refract at 90° at critical angle.By Snell's law, 2×sin37°=μ(x)×sin90°μ(x)=2×53=56
When the ray travels from top to bottom of the slab,the refractive index of slab decreases (μ(x)=a−bx) and the ray will refract at 90° at critical angle.By Snell's law, 2×sin37°=μ(x)×sin90°μ(x)=2×53=56a−bx=56
When the ray travels from top to bottom of the slab,the refractive index of slab decreases (μ(x)=a−bx) and the ray will refract at 90° at critical angle.By Snell's law, 2×sin37°=μ(x)×sin90°μ(x)=2×53=56a−bx=562−x=56
When the ray travels from top to bottom of the slab,the refractive index of slab decreases (μ(x)=a−bx) and the ray will refract at 90° at critical angle.By Snell's law, 2×sin37°=μ(x)×sin90°μ(x)=2×53=56a−bx=562−x=56x=54m
Explanation:
Applying snell’s 1aω :
n
1
sinγ
1
=n
2
sinγ
2
⇒n
1
sinθ=n
2
sinγ
2
⇒γ
2
=sin
−1
(
n
2
n
1
sinθ)
γ
2
+θ+120
∘
=180
∘
(according to question,)
⇒γ
2
=60
∘
−θ
⇒sin
−1
(
n
2
n
1
sinθ)= 60
∘
−θ
⇒
n
2
n
1
sinθ=sin(60
∘
−θ)
⇒
n
2
n
1
sinθ=sin60
∘
cosθ−cos60
∘
sinθ
⇒
n
2
n
1
=
2
3
×
tanθ
1
−
2
1
⇒
n
2
n
1
=
2+anθ
3
−tanθ
⇒
n
1
n
2
=
3
−tanθ
2tanθ
. sin
−1
(
n
1
n
2
)=sin
−1
(
3
−tanθ
2tanθ
)
Lritical angle =sin
−1
(
3
−tanθ
2tanθ
)