A beam on light falls on the glass (n_2 = 1.4) of thickness 10 cm from the air (

Question

A beam on light falls on the glass (n_2 = 1.4) of thickness 10 cm from the air (n_i = 1) at angle theta_i = 30 degree (see Fig.). After the ray goes through the glass it enters oil (n_3 = 1.2). Find angle of the ray in the oil. How far away horizontally the ray' will move in the glass? theta = 24.6 degree x = 3.8 cm An optical fiber has index of refraction n = 1.5. If outside medium is air, what is the critical angle for total internal reflection? For a light ray entering the fiber what is the largest angle with the axis of the fiber at which total internal refection is possible? If we put our fiber in the water (n_w = 1.3) is this angle (with the fiber axis) going to be larger or smaller? Theta_c = 41.8 degree, theta_enter = 48.2 degree, smaller

Explanation / Answer

2)

let theta_1i and theta_1r are the angle of incidence and and angle of refraction at first interface.

sin(theta_i1)/sin(theta_r1) = n2/n1

sin(30)/sin(theta_r1) = 1.4/1

sin(theta_r1) = sin(30)/1.4

theta_r1 = sin^-1(0.357)

= 20.92 degrees

at glass-oil interface, angle of incidense, theta_2i = 20.92 degrees

theta_2r = ?

again apply,

sin(theta_2i)/sin(theta_2r) = n3/n2

sin(20.92)/sin(theta_2r) = 1.2/1.4

sin(theta_2r) = (1.4/1.2)*sin(20.92)

theta_2*r = sin^-1(0.417)

= 24.6 degrees <<<<<<<<<<<-----------------Answer

let x is the horizontal distance travelled by the ray

now apply, tan(theta_1r) = x/d

==> x = d*tan(theta_1r)

= 10*tan(20.92)

= 3.82 cm <<<<<<<<<<<-----------------Answer

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