Solving Refraction Problem: Aim Flashlight at What Angle?

  • Thread starter roxxyroxx
  • Start date
  • Tags
    Refraction
In summary, the problem involves a penny located 0.75m from the edge of a pool that is 1.5m deep. A flashlight is used to illuminate the penny from the edge of the pool. Using the index of refraction for water, the flashlight should be aimed at an angle of 54 degrees to the pool deck in order to properly illuminate the penny. The solution involves using Snell's Law and the sin/sin rule.
  • #1
roxxyroxx
48
0

Homework Statement



A penny lies on the bottom of a pool 0.75m from the edge of the pool and 1.5m below the surface. A flashlight beam is shone over the edge of the pool to illuminate the penny. At what angle to the pool deck should the flashlight be aimed?

Homework Equations



??

The Attempt at a Solution


??
 
Physics news on Phys.org
  • #2
roxxyroxx said:

Homework Statement



A penny lies on the bottom of a pool 0.75m from the edge of the pool and 1.5m below the surface. A flashlight beam is shone over the edge of the pool to illuminate the penny. At what angle to the pool deck should the flashlight be aimed?

You're going to need the index of refraction aren't you?

Water is 1.33. Is that what you are supposed to use?

If you don't know it already, then maybe better look at Snell's Law.
 
  • #3
Hi roxxyroxx! :wink:
roxxyroxx said:
A penny lies on the bottom of a pool 0.75m from the edge of the pool and 1.5m below the surface. A flashlight beam is shone over the edge of the pool to illuminate the penny. At what angle to the pool deck should the flashlight be aimed?

(I assume it means that the light enters the pool exactly at the edge?)

Use the sin/sin rule …

show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
 
  • #4
ok well i did sin(0.75/1.5) --> sin^-1(0.75/1.5) giving me 30 degrees so:
(1.00)(sin 30) = (1.33)(sin theta)
sin theta = 0.3759...
theta = 22 degrees
but the answer should be 54 degrees
 
  • #5
roxxyroxx said:
ok well i did sin(0.75/1.5) --> sin^-1(0.75/1.5) giving me 30 degrees so:
(1.00)(sin 30) = (1.33)(sin theta)
sin theta = 0.3759...
theta = 22 degrees
but the answer should be 54 degrees

Isn't the angle you want from entering the water given by

tan-1(.75/1.5) = θ
 
  • #6
ok so tan^-1(0.75/1.5) = 26.565...
(1.33)(sin 26.565...) = (1.00)(sin theta)
sin theta = 0.595
theta = 37 degrees
but the answer is 54 ..
 
  • #7
roxxyroxx said:
ok so tan^-1(0.75/1.5) = 26.565...
(1.33)(sin 26.565...) = (1.00)(sin theta)
sin theta = 0.595
theta = 37 degrees
but the answer is 54 ..
This angle is the angle which the light makes with the normal to the pool. But the answer required is the angle to the pool deck.
So the required angle is 90 - theta
 
  • #8
ookk thank you! >.<
 

Related to Solving Refraction Problem: Aim Flashlight at What Angle?

What is refraction and why is it important?

Refraction is the bending of light as it passes through different mediums. It is important because it helps us understand how light behaves and how it can be manipulated to create different effects. It also plays a critical role in various scientific and technological applications, such as in lenses and optical instruments.

How do I calculate the angle of refraction?

The angle of refraction can be calculated using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speed of light in the first medium to the speed of light in the second medium. This can be represented by the equation: n1sinθ1 = n2sinθ2, where n1 and n2 are the refractive indices of the two mediums.

How do I solve a refraction problem when using a flashlight?

To solve a refraction problem involving a flashlight, you need to know the angle of incidence, the refractive index of the medium the light is passing through, and the angle of refraction. You can then use Snell's law to calculate the missing angle or refractive index. It is also important to keep in mind that the angle of incidence and the angle of refraction are measured with respect to the normal, which is a perpendicular line to the surface the light is passing through.

What factors affect the angle of refraction?

The angle of refraction is affected by the difference in the refractive indices of the two mediums, the angle of incidence, and the wavelength of light. As the angle of incidence increases, the angle of refraction also increases. Additionally, different materials have different refractive indices, which can cause the angle of refraction to vary. The wavelength of light also plays a role, as shorter wavelengths experience more refraction than longer wavelengths.

How can I use refraction to my advantage?

Refraction can be used in various ways, such as in the design of lenses for eyeglasses and cameras, in the creation of optical illusions, and in the production of fiber optics for communication. It can also be used to analyze the composition of materials by measuring the angle of refraction, as different materials have different refractive indices. By understanding the principles of refraction, scientists and engineers can create new technologies and improve existing ones.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
434
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top