# A Maximum angle of acceptance - optical fibre

1. Aug 20, 2016

### Petra de Ruyter

Hi all

Tearing my hair out trying to calculate the Maximum acceptance angle for a problem. Experiment involved shining a laser into a tube of water. I have calculated the angle of refraction based on the average of two critical angles (one known and one measured manually) as 52.49 degrees. Asked to calculate the maximum angle of acceptance with just the critical and the angle of refraction. Im of the opinion that more information is needed, can someone please clarify.

Ive tried several geometric ways around this and can't get it?
Cheers

2. Aug 20, 2016

### blue_leaf77

So, if I am not mistaken the water acts as the core and the glass as cladding? However you know that the core must have higher refractive index than cladding which is hardly achievable for such configuration (most glasses have higher refractive index than water).
Between which media did you get this value?

3. Aug 20, 2016

### Petra de Ruyter

Air and water :-)

4. Aug 20, 2016

### blue_leaf77

The critical angle between water and air is about $48.6^o$, I wonder what are the known and measured values of critical angle with which you obtained that averaged value. Can you tell us?

5. Aug 21, 2016

### Petra de Ruyter

53.00 and 51.98 The laser that was provided was pretty weak and the angles are a bit sketchy.

Cheers

6. Aug 21, 2016

### blue_leaf77

Regardless of that peculiarity of the given value, the acceptance angle is calculated using ray optics. The ray diagram is illustrated in the picture below.

You want to express $\theta_a$ in terms of $\theta_c$. Clearly you need to know $n_i$, $n_f$, and $n_c$.

7. Aug 22, 2016

### Khashishi

What do you mean by critical angle?

8. Aug 22, 2016

### sophiecentaur

9. Aug 23, 2016

### Khashishi

Maybe I should have said, how are you measuring the critical angle? And why can't you measure the acceptance angle?

10. Aug 24, 2016

### Petra de Ruyter

Hi there. I could not measure it because I did not have the refractive index, now I do and all is well. many thanks for all your assistance.