Optics - Refraction and a transparent sphere

[Saitama]

Optics -- Refraction and a transparent sphere

Homework Statement

A spider is hanging by means of its own silk thread directly above a transparent fixed sphere of radius R=20 cm as shown in the figure. The refractive index of the material of the sphere is equal to $\sqrt{2}$ and the height of the spider from the centre of the sphere is 2R. An insect, initially sitting at the bottom, starts crawling on the sphere along a vertical circular path with the constant speed of $\frac{\pi}{4}$ cm/s. For how long time the insect will be invisible for the spider, assume that it crawls once round the vertical circle.
(see attachment)

Homework Equations

The Attempt at a Solution

Honestly, I have no idea. I don't have a clue about which equations I have to use here.

Any help is appreciated. Thanks!

 

[SammyS]

Homework Statement

A spider is hanging by means of its own silk thread directly above a transparent fixed sphere of radius R=20 cm as shown in the figure. The refractive index of the material of the sphere is equal to $\sqrt{2}$ and the height of the spider from the centre of the sphere is 2R. An insect, initially sitting at the bottom, starts crawling on the sphere along a vertical circular path with the constant speed of $\frac{\pi}{4}$ cm/s. For how long time the insect will be invisible for the spider, assume that it crawls once round the vertical circle.
(see attachment)

Homework Equations

The Attempt at a Solution

Honestly, I have no idea. I don't have a clue about which equations I have to use here.

Any help is appreciated. Thanks!

This question has to do with total internal reflection. Start there.

 

[Saitama]

This question has to do with total internal reflection. Start there.

I did think of this before but the insect keeps on moving which is confusing me.

 

[SammyS]

I did think of this before but the insect keeps on moving which is confusing me.

Find out the location of the insect just as a ray from the insect to the spider reaches the point of criticality.

I actually found this easier to do by tracing the ray from the spider, back to the insect.

 

[Saitama]

I actually found this easier to do by tracing the ray from the spider, back to the insect.

I too think this would be easier.
If a ray from the spider is tangent to the sphere, then by Snell's law, it deviates towards the normal by pi/4. (see attachment)

I am facing problems with geometry. Look at the triangle where the angles come out to be 150 and 45 degrees, the sum goes over 180 degrees! I can't find where I went wrong.

EDIT: Woops, please ignore this reply. I think I have found out my mistake. One more question, would the rays meet, after refraction, at the point where insect was initially placed?

 

[SammyS]

I too think this would be easier.
If a ray from the spider is tangent to the sphere, then by Snell's law, it deviates towards the normal by pi/4. (see attachment)

I am facing problems with geometry. Look at the triangle where the angles come out to be 150 and 45 degrees, the sum goes over 180 degrees! I can't find where I went wrong.

EDIT: Woops, please ignore this reply. I think I have found out my mistake. One more question, would the rays meet, after refraction, at the point where insect was initially placed?

They don't meet there, if my solution is correct.

The angle you have labelled as 60° is 120° .

 

[Saitama]

The angle you have labelled as 60° is 120° .

Yes, I figured that out already, that's why I said to ignore the reply. I will be back with a solution.

 

[SammyS]

Yes, I figured that out already, that's why I said to ignore the reply. I will be back with a solution.

After being refracted, the ray from the spider then intercepts the circle again. A radius from each place the ray intercepts the circle together with the ray itself, form an isosceles right triangle, with the ray falling along the hypotenuse.

 

[Saitama]

I am still not getting the right answer.
(see attachment)
The angular displacement of the insect when it is invisible to the spider is $2\cdot \frac{13\pi}{24}$.
The time taken is $t=\frac{\theta}{\omega}$ where $\omega$ is the angular velocity of insect.
Substituting the values, $t=\frac{260}{3} sec$ which is wrong. :(

 

[SammyS]

Your drawings are misleading.

The refracted ray makes an angle of 45° (π/2) with respect to the normal .

In both drawings, the refracted ray appears to be drawn at an angle of 45° with respect to the horizontal.

 

[Saitama]

The spider is visible while it crawls along the red arc. What central angle belongs to it?

ehild

How do you know that the rays go like you have shown in your diagram?

 

[ehild]

The bug is visible along the red arc. The refractive index is 1/√2 so the angle of refraction is 45°. The radii make 90°angle between the point of incidence and point where the refracted ray arrives to the circle again.

ehild

 

[Saitama]

Thank you both for your inputs, I have got the right answer!

 

[SammyS]

How do you know that the rays go like you have shown in your diagram?

You, yourself, said the angle of refraction is π/4 .

 

[Saitama]

You, yourself, said the angle of refraction is π/4 .

Yes, I did but I was confused by the path the rays would take after refraction. I messed up with simple geometry.

 

[ehild]

Yes, I did but I was confused by the path the rays would take after refraction. I messed up with simple geometry.

The rays travel along straight lines

 

[Saitama]

The rays travel along straight lines

Yes but see my previous diagrams, they are completely wrong.

 

[ehild]

Yes, the upper central angle was wrong, and also, you did not draw the refracted light according to the angle you got.

ehild.