Answer:
The angle of refraction of sheet 3 when sheet 1 is on top of it is [tex]\theta_{r_s } = 23.1 ^o[/tex]
Explanation:
From the question we are told that
The angle of incidence is [tex]\theta _i = 26.50 ^o[/tex]
The angle of refraction angle for sheet 1 is [tex]\theta _{r_1}} = 31.70 ^o[/tex]
The angle of refraction for sheet 3 is [tex]\theta _{r_3}} = 36.70 ^o[/tex]
According to Snell's law
[tex]\frac{n_2}{n_1} = \frac{sin (\theta_1)}{sin (\theta_{r_1})}[/tex]
Where [tex]n_1 \ and \ n_2[/tex] are refractive index of sheet 1 and sheet 2
=> [tex]n_2 = n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})}[/tex]
Also when sheet 3 in on top of sheet 2
[tex]\frac{n_2}{n_3} = \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
substituting for [tex]n_2[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
=> [tex]n_3 = n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}[/tex]
when sheet 1 in on top of sheet 3
[tex]\frac{n_3}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
where [tex]r_s[/tex] is the angle of refraction when sheet 1 is on top of sheet 3
substituting for [tex]n_3[/tex]
[tex]\frac{ n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
=> [tex]sin (\theta _{r_s}) = n_1 * sin (\theta_i) * \frac{sin (\theta_{r_1})}{ n_1 * sin(\theta_{r_3})}[/tex]
substituting values
[tex]sin (\theta _{r_s}) = n_1 * sin (26.50) * \frac{sin (31.70)}{ n_1 * sin(36.70)}[/tex]
=> [tex]\theta_{r_s } = sin^{-1} (0.3923)[/tex]
=> [tex]\theta_{r_s } = 23.1 ^o[/tex]
