Respuesta :
Answer:
Angle of transmitted ray is [tex]29.14^{o}[/tex]
Explanation:
According to snell's law we have
[tex]n_{1}sin(\theta _{i})=n_{2}sin(\theta r)[/tex]
Since the incident medium is air thus we have [tex]n_{1}=1[/tex]
By definition of refractive index we have
[tex]n=\frac{c}{v}[/tex]
c = speed of light in vacuum
v = speed of light in medium
Applying values we get
[tex]n_{2}=\frac{3\times 10^{8}}{2.5\times 10^{8}}=1.2[/tex]
Thus using the calculated values in Snell's law we obtain
[tex]sin(\theta _{r})=\frac{sin(31.3}{1.2}\\\\\therefore sin(\theta _{r})=0.4329\\\\\theta_{r}=sin^{-1}(0.4329)\\\\\theta _{r}=29.14[/tex]
Answer:
Angle made by the transmitted ray = 25.65°
Explanation:
Speed of light in plastic = v = 2.5 × 10⁸ m/s
refractive index of plastic (n₂) / refractive index of air (n₁)
= speed of light in air c / speed of light in plastic v.
⇒ n₂ = (3× 10⁸) / (2.5 × 10⁸) = 1.2
Angle of incidence = 31.3° = i
n₁ sin i = n₂ sin r
⇒ sin r = (1)(0.5195) / 1.2 = 0.4329
⇒ Angle made by the transmitted ray = r = sin⁻¹ (0.4329) =25.65°
