Light traveling through air is incident on a flat piece of...

Name: light traveling through air is incident on a flat piece of glass at a 35circ angle of incidence and enters the glass at an angle of refraction thetaglass suppose that a layer of water is ad 28015
Uploaded: 2021-12-14T15:12:31-08:00
Duration: 6 min 54 s
Channel: Aparna Shakti
Description: light traveling through air is incident on a flat piece of glass at a 35circ angle of incidence and enters the glass at an angle of refraction thetaglass suppose that a layer of water is ad 28015

Light traveling through air is incident on a flat piece of glass at a $35^{\circ}$ angle of incidence and enters the glass at an angle of refraction $\theta_{glass}$. Suppose that a layer of water is added on top of the glass. Then the light travels through air and is incident on the water at the $35^{\circ}$ angle of incidence. Does the light enter the glass at the same angle of refraction $\theta_{glass}$ as it did when the water was not present? Yes No

Transcript

00:01 In this problem, a thin frame of water sits on top of flat piece of glass and light is incident from below figure.

00:14 If the angle of incidence is 30 degree, what is the angle of refraction of the final outgoing rate, right? so this is how it is given.

00:27 So this is one like this.

00:29 Okay so this is going like this okay so this is the layer of you can see this was the layer of glass right and then it is water right so here you can see that refractive index for water is 1 .33 and refractive index of glass is 1 .55 and this is air okay so now this is refractive so first of all with glass, it will be 30 degree making an, you can say an angle 30 degree, right? then it is getting refracted like this, then like this, let us say, right? something slant.

01:25 So now suppose this is a normal, okay? this is a normal.

01:35 All right.

01:35 So that means this angle will be, if this angle is making with 1 .5.

01:40 Vertical right that is 30 degrees so this should be 30 degree isn't it so so this is interface of medium right these are mediums glass water and air are medium and light is getting refracted here right so this is medium one this is two and this is three okay so now um let us say from snail's law what we know we know that n1 sign 3.

02:18 Theta 1 equals to n2 sine theta 2 right so n is refractive index remember this okay so for the interface of medium glass and water what we can write here we can write it as n1 is and 1 is you can say 1 .55 right 5 5 right 1 .5 sign okay so let us don't write it as okay so this is something else okay so this is not vertically opposite because this is not a straight line remember so this has not to be equals to this okay so let me write it as theta 2 right and this is theta 1 why because suppose uh this is a line and it is an angle and this is making an angle if this is a straight line then only this vertically opposite angle are equal right otherwise they will not be equal so here there is real fraction occurring so that's why the angle will be something else right more than 30 degree it seems like okay so now um sine 30 will be equals to 1 .55 into sine 30 that is given by okay 1 .33 sine theta 2 right so when we will calculate it we will be getting theta 2 equals to 48 point 5 .9 degree.

04:18 Right? so this is the answer...

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