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?

Explanation:

Formula of lateral displacement

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°

= .45 x .454 / .9455

= .216 m

= 21.6 cm .

c. magnitude and direction

where is the photo and the answer choices?