Diameter of circle of light seen from above water - illuminated beneath

[Marcargo]

Homework Statement

A small light is 22.0 cm below the surface of a liquid of refractive index 1.50. Viewed from above, the light appears to illuminate a circle on the surface of the water. What is the diameter of the circle?
cm

Homework Equations

Snell's Law
n₁sin(θ)₁ = n₂sin(θ)₂

The Attempt at a Solution

Based on Snell's law:
1.6 x 1 = 1 x sinθ, if I can find the angle of refraction, then maybe I can find the diameter of the circle with trigonometry?
I can't really see a way through this one - I'm thinking that I want to find the 'radius' of the circle, and from there, find the diameter. Maybe the distance that the light is used below the circle can be then used as a side for trigonometry to find the radius?

 

[Curious3141]

Homework Statement

A small light is 22.0 cm below the surface of a liquid of refractive index 1.50. Viewed from above, the light appears to illuminate a circle on the surface of the water. What is the diameter of the circle?
cm

Homework Equations

Snell's Law
n₁sin(θ)₁ = n₂sin(θ)₂

The Attempt at a Solution

Based on Snell's law:
1.6 x 1 = 1 x sinθ, if I can find the angle of refraction, then maybe I can find the diameter of the circle with trigonometry?
I can't really see a way through this one - I'm thinking that I want to find the 'radius' of the circle, and from there, find the diameter. Maybe the distance that the light is used below the circle can be then used as a side for trigonometry to find the radius?

The problem is not quite complete. They did not specify the geometry of the light source. However, the usual simplifying assumption is that of a point source of monochromatic light that emits isotropically (i.e. same intensity in all directions).

What you should be thinking of is the phenomenon of total internal reflection when the angle of incidence exceeds the critical angle. Think about this: light rays emerge at all angles from the light source. Does every ray that hits the surface actually escape into the air? Can you construct a "limiting cone" for the light that actually escapes the water?

 

[tiny-tim]

Welcome to PF!

Hi Marcargo! Welcome to PF!

Based on Snell's law:
1.6 x 1 = 1 x sinθ, if I can find the angle of refraction, then maybe I can find the diameter of the circle with trigonometry?

(what's 1.6? )

yes, you've used θ₁ = 90°, so that gives you θ₂

so you have a right-angled triangle, and you know the angle and one side …

where's the difficulty? ​

 

[ehild]

Based on Snell's law:
1.6 x 1 = 1 x sinθ, if I can find the angle of refraction, then maybe I can find the diameter of the circle with trigonometry?

(what's 1.6? )

yes, you've used θ₁ = 90°, so that gives you θ₂

so you have a right-angled triangle, and you know the angle and one side …

where's the difficulty? ​

The difficulty is that "1.6 x 1 = 1 x sinθ" produces sin(θ)=1.6 .

ehild

 

[Marcargo]

Ok, so I've worked out the critical angle for the light - sin^-1(n₂/n₁)
= 41.8°

This gives me the angle, and from trig, I used tan θ = r/30
= 26.8cm

diameter = 26.8 x 2 = 54cm (2sf)


Thank you for the help everyone! It actually feels really great to have found this physics community, makes be love the subject even more :)


Marcargo.

 

[gneill]

Ok, so I've worked out the critical angle for the light - sin^-1(n₂/n₁)
= 41.8°

This gives me the angle, and from trig, I used tan θ = r/30
= 26.8cm

diameter = 26.8 x 2 = 54cm (2sf)

Where did the 30 come from?

 

[tiny-tim]

Where did the 30 come from?

he keeps changing the question!

A small light is 22.0 cm below the surface of a liquid of refractive index 1.50.

This gives me the angle, and from trig, I used tan θ = r/30
= 26.8cm

diameter = 26.8 x 2 = 54cm (2sf)

shouldn't it be three significant figures ?

 

[Curious3141]

Ok, so I've worked out the critical angle for the light - sin^-1(n₂/n₁)
= 41.8°

This gives me the angle, and from trig, I used tan θ = r/30
= 26.8cm

diameter = 26.8 x 2 = 54cm (2sf)


Thank you for the help everyone! It actually feels really great to have found this physics community, makes be love the subject even more :)


Marcargo.

I have no idea where the 30 came from, but if the original question is taken as accurate (22cm), I get a different answer.

There is no need to do the arcsin and then take the tangent, you can use simple trig identities to manipulate it algebraically to get an exact answer of $\frac{88\sqrt{5}}{5}$, which can then be evaluated by calculator.