Respuesta :
Answer:
Explanation:
Formula of lateral displacement
[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]
t is thickness of slab , i and r are angle of incidence and refraction respectively .
Given t = .45 m
sin i / sin r = 2.2
sin 46 / sin r = 2.2
sin r = .719 / 2.2 = .327
r = 19°
[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]
[tex]S_{lateral}=\frac{.45}{cos19} \times sin(46-19)[/tex]
= .45 x .454 / .9455
= .216 m
= 21.6 cm .
The displacement, D, of the beam when it exits the slab is; 21.65 cm.
We are given;
Refractive index of slab material; nm = 2.2
Thickness of slab; t = 0.45 m
Refractive index of air; na = 1
Angle of incidence; θa = 46°
From snell's law, we can calculate the angle of refraction from;
na × sin θa = nm × sin θm
Thus;
1 × sin 46 = 2.2 × sin θm
0.7193 = 2.2 × sin θm
sin θm = 0.7193/2.2
θm = sin^(-1) 0.32695
θm = 19.08°
Formula for the displacement of the beam is;
D = (t/cos θm) × sin (θa - θm)
Plugging in the relevant values gives;
D = (0.45/cos 19.08) × sin (46 - 19.08)
D = 0.4783 × 0.4527
D = 0.2165m = 21.65 cm
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