Optics -- Refraction and a transparent sphere

Pranav-Arora

1. The problem statement, all variables and given/known data
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)

2. Relevant equations



3. 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!

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SammyS

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

Pranav-Arora

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

SammyS

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.

Pranav-Arora

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?

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SammyS

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



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

Pranav-Arora

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

SammyS

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.

Pranav-Arora

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. :(

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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.

Pranav-Arora

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

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Pranav-Arora

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

SammyS

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

Pranav-Arora

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

ehild

The rays travel along straight lines

Pranav-Arora

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