Light passes from air into flint glass at a nonzero angle of...

Name: Light passes from air into flint glass at a nonzero angle of incidence. (a) Is it possible for the component of its velocity perpendicular to the interface to remain constant? Explain your answer. (b) What If? Can the component of velocity parallel to the interface remain constant during refraction? Explain your answer.
Uploaded: 2022-03-07T13:51:54-08:00
Duration: 8 min 6 s
Channel: Vishal Gupta
Description: Light passes from air into flint glass at a nonzero angle of incidence. (a) Is it possible for the component of its velocity perpendicular to the interface to remain constant? Explain your answer. (b) What If? Can the component of velocity parallel to the interface remain constant during refraction? Explain your answer.

Light passes from air into flint glass at a nonzero angle of incidence. (a) Is it possible for the component of its velocity perpendicular to the interface to remain constant? Explain your answer. (b) What If? Can the component of velocity parallel to the interface remain constant during refraction? Explain your answer.

Transcript

00:02 Hi, here in this given problem this is the interface separating air from flint glass.

00:17 Then there is a normal drone at the interface and a ray of light passing from air to flint glass.

00:28 This is angle of incidence i in the air.

00:34 This is angle of refraction r in the flint glass.

00:40 It is given that index of refraction for the glass, flint glass, this is 1 .66.

00:50 So using an expression for index of refraction as c by v, where c is the speed of light in the air, and v is the speed of light in the glass.

01:05 So we can say this is 1 .66.

01:08 Hence, we get an expression for the speed of light in the flint glass and that is c by 1 .66 means clearly speed of light will be less in flint glass as compared with that in the air.

01:26 Now, in the first part of the problem, we take component of the velocity of the light perpendicular to the surface.

01:39 So here this component this will be c c c i and here this component in the glass perpendicular to the surface.

01:53 This will be v c c r.

02:00 Now as this length of the component increases, v is decreasing no doubt but r is also decreasing so cos r will be increasing so we can say as i is more than r so cos i will be less than cos r and v is less than c but it is possible that c c c c may be equal to v.

02:52 Coss r which is the given condition and this is possible also.

02:56 We make it equation number one.

02:59 Then using snell's law which says sine i by sine r is equal to index of refraction which is 1 .66 hence sine i is equal to 1 .66 sine r.

03:28 We can consider it to be equational 2.

03:31 Then multiplying equation 1 by equation 2, left -hand side with the left -hand side.

03:36 So we get c into sine i into cos i is equal to 1 .66v into sine r.

03:49 1 into cos r.

03:51 Now if we multiply both sides by 2, we get c into 2 sine i into cosi is equal to 1 .66v again 2 to sine r into cos r...