The smallest angle flashlight beam can make with horizontal

Click For Summary

Homework Help Overview

The problem involves determining the smallest angle a flashlight beam can make with the horizontal while illuminating the bottom of a cylindrical tank. The tank has specific dimensions, and the refractive index of water is provided, though its relevance is questioned.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the geometry of the situation, including the assumption that the light must reflect off the side of the tank. Questions arise about the validity of this assumption and the implications for determining the angle.

Discussion Status

There is ongoing exploration of the problem setup, with participants questioning the assumptions made about the light's path and the angle needed to illuminate the bottom. Some guidance has been offered regarding the need to visualize the tank's cross-section and reconsider the light's trajectory without reflections.

Contextual Notes

Participants express uncertainty about the approach due to potential misunderstandings of the problem setup and the information provided. There is a suggestion to draw a cross-section of the tank to clarify the light's path.

armxneo
Messages
3
Reaction score
0

Homework Statement


At the aquarium where you work, a fish has gone missing in a 10.5m -deep, 9.45m -diameter cylindrical tank. You shine a flashlight in from the top edge of the tank, hoping to see if the missing fish is on the bottom. What’s the smallest angle your flashlight beam can make with the horizontal if it’s to illuminate the bottom?

nwater=1.33 (Not sure if this is relevant)

Homework Equations



tan(θ)=y/x
n1sinΘ=n2cosθ

The Attempt at a Solution



I first drew a picture of a cylinder with height 10.5 meters and 9.45 meters
I assume that in order to illuminate the bottom of the tank, the light needs to hit the side of the tank at half of it's height (if this is incorrect, then I am not sure how to determine where the light hits the side of the tank)

At height 10.5/2= 5.25 meters, I solved for Θ; tan(Θ)=5.25/9.45
tan-1(5.25/9.45)= 29.05 degrees

I used this equation in snell's law:
1.33sin(29.05)=(1)sin(θair)
θ=40.23.
I found the θ to the normal, so to find θ to the horizontal: 90-40.23= 49.77 degrees.

This answer is incorrect, so any help would be appreciated!
 
Physics news on Phys.org
Why at half its height? Are you assuming reflection off the side? I would not assume that.
 
I know it reflects off the side because this would give a smaller angle to the horizontal (which is what the question is asking me to optimize for). I am unsure about how I would find an angle, and where the light would reflect off of if it was not at half the height.
 
armxneo said:
I know it reflects off the side because this would give a smaller angle to the horizontal (which is what the question is asking me to optimize for).
That does not follow. Suppose it does not reflect. It is possible to hold the light so that it illuminates (some part of) the bottom; at a vey low angle it will not iluminate the bottom; therefore there is a least angle at which it illuminates the bottom.
Conversely, if it can reflect off one side, why not both sides?
 
Ok. Based on that I have no idea how to approach the problem with the given information.
 
armxneo said:
Ok. Based on that I have no idea how to approach the problem with the given information.
This makes me think you have misunderstood the set up, but I can't guess in what way. Draw a cross section of the tank. It's a rectangle. The source of light is at top left (say). To just reach the bottom, without reflections, at the lowest possible angle, where will the light beam strike the bottom?
 

Similar threads

Fish has gone missing in a tank
  • · Replies 6 ·
Replies
6
Views
4K