Question

A flat slab of material (nm = 2.2) is d = 0.45 m thick. A beam of light in air (na = 1) is incident on the material with an angle θa = 46 degrees with respect to the surface's normal.
Numerically, what is the displacement, D, of the beam when it exits the slab?

Answer 1

Answer:

Explanation:

Formula of lateral displacement

$S_{lateral}=\frac{t}{cosr} \times sin(i-r)$

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°

$S_{lateral}=\frac{t}{cosr} \times sin(i-r)$

$S_{lateral}=\frac{.45}{cos19} \times sin(46-19)$

= .45 x .454 / .9455

= .216 m

= 21.6 cm .