Respuesta :
Answer:
r = 71.8⁰
Explanation:
given,
refractive index of the glass 1 = 1.70
refractive index of glass 2 = 1.58
angle of incidence = 62°
angle of refraction =?
using Snell's law
[tex]\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}[/tex]
[tex]\dfrac{sin\ 62^0}{sin\ r} = \dfrac{1.58}{1.70}[/tex]
1.7 ×sin 62 ^0 = 1.58× sin r
[tex]sinr = \dfrac{1.7\times sin 62^0}{1.58}[/tex]
sin r = 0.95
r = sin⁻¹(0.95)
r = 71.8⁰
angle of refraction =r = 71.8⁰
Answer:
Angle of refraction, [tex]\theta_{r} = 71.804^{\circ}[/tex]
Solution:
As per the question:
Refractive index of the first sheet, [tex]mu_{a} = 1.70[/tex]
Refractive index of the second sheet, [tex]mu_{b} = 1.58[/tex]
Angle of incidence, [tex]\theta_{i} = 62.0^{\circ}[/tex]
Now, according to Snell's Law:
[tex]\mu_{a}sin\theta_{i} = \mu_{b}sin\theta_{r}[/tex] (1)
[tex]\theta_{r}[/tex] = Angle of refraction
From eqn(1), angle of refraction may be computed as:
[tex]sin\theta_{r} = \frac{\mu_{a}}{\mu_{b}}sin\theta_{\theta_{i}}[/tex]
[tex]sin\theta_{r} = \frac{1.70}{1.58}sin(62^{\circ})[/tex]
[tex]\theta_{r} = sin^{- 1}(0.9499) = 71.804^{\circ}[/tex]
