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Question Seven – 12 Marks a) In a simple pipe (fig 5), a ray of light must be transmitted, with minimum loss, along a core. List three major conditions, for this to be possible. [3 Marks] axis 0° external medium $n_3$ incident light cladding $n_2$ core $n_1$ fig 5 b) The diagram shows a ray light, entering the surface of oil at an unknown angle of incidence $\theta_0$. The light passes through the oil, strikes and enters the glass and is just totally internally reflected at the glass/air boundary. i. Calculate the critical angle at the glass/air boundary. [2 Marks] ii. Hence determine the path the light takes, as it passes through the different media. Calculate the angles of incident and refraction at each surface. [7 Marks] $n_{glass} = 1.65$ $65^\circ$ air air $\theta_c$ $n_{oil} = 1.45$ Incident ray

          Question Seven – 12 Marks
a) In a simple pipe (fig 5), a ray of light must be transmitted, with minimum loss, along a core. List three major conditions, for this to be possible. [3 Marks]
axis
0°
external medium $n_3$
incident light
cladding $n_2$
core $n_1$
fig 5
b) The diagram shows a ray light, entering the surface of oil at an unknown angle of incidence $\theta_0$. The light passes through the oil, strikes and enters the glass and is just totally internally reflected at the glass/air boundary.
i. Calculate the critical angle at the glass/air boundary. [2 Marks]
ii. Hence determine the path the light takes, as it passes through the different media. Calculate the angles of incident and refraction at each surface. [7 Marks]
$n_{glass} = 1.65$
$65^\circ$
air
air
$\theta_c$
$n_{oil} = 1.45$
Incident ray

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question seven 12 marks a in a simple pipe fig 5 a ray of light must be transmitted with minimum loss along a core list three major conditions for this to be possible 3 marks axis 0 external 97961

Added by Mackenzie N.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Vishal Gupta

03/18/2022

Step 1

It can be calculated using Snell's Law: $n_1 \sin \theta_1 = n_2 \sin \theta_2$. In this case, $n_1$ is the refractive index of glass ($n_{glass} = 1.65$), $n_2$ is the refractive index of air ($n_{air} = 1$), $\theta_1 = \theta_c$, and $\theta_2 = 90^\circ$. Show more…

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Question Seven – 12 Marks a) In a simple pipe (fig 5), a ray of light must be transmitted, with minimum loss, along a core. List three major conditions, for this to be possible. [3 Marks] axis 0° external medium $n_3$ incident light cladding $n_2$ core $n_1$ fig 5 b) The diagram shows a ray light, entering the surface of oil at an unknown angle of incidence $\theta_0$. The light passes through the oil, strikes and enters the glass and is just totally internally reflected at the glass/air boundary. i. Calculate the critical angle at the glass/air boundary. [2 Marks] ii. Hence determine the path the light takes, as it passes through the different media. Calculate the angles of incident and refraction at each surface. [7 Marks] $n_{glass} = 1.65$ $65^\circ$ air air $\theta_c$ $n_{oil} = 1.45$ Incident ray

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Transcript

0:00 Hi.

00:02 Here in this given problem, first of all there is a boundary separating medium number 1 from medium number 2.

00:16 Then one more boundary separating again this medium number 2 from medium number 1 again.

00:25 It is repeated.

00:26 Array of light is entering while traveling in medium number 1 enters into this medium number 2 making an angle of incidence here, theta i.

00:46 Then it makes an angle of transmission here in this medium number 2, that is theta t.

00:57 The same angle will be serving as angle of incidence for the refraction of this ray of light again from medium number 2 to the medium number 1.

01:12 So angle of refraction here, sorry, angle of incidence here, that is also theta t.

01:22 And finally the ray of light will come back into the same medium, making this angle of refraction theta f we can say it to be.

01:38 Now, n1 means the index of refraction of medium 1.

01:44 This is given as 1 .05, n2, index of refraction of medium 2.

01:52 This is 1 .40.

01:55 Angle of incidence in the first medium, this is given as 60 .0 degree.

02:00 In the first part of the problem, we have to find angle of transmission, theta t.

02:05 So using snell's law for the refraction from medium 1 to medium 2 it is given as index of refraction of the second medium with respect to the first one that is equal to sine theta i divided by sine theta t plugging in all the known values this is 1 .40 divided by 1 .05 is equal to sine 60 .0 degree and divided by sine theta t.

02:46 So we get an expression for this sine theta t and this is equal to sine 60 degree multiplied by 1 .05 divided by 1 .40 which comes out to be equal to 0 .6 so, finally, theta t here will be given by sine inverse of 0 .6495 means theta t is 40 .5 degree answer for the first part of the problem here...

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