Calculating Velocity of Reflected Image in Concave Mirror

Click For Summary

Homework Help Overview

The problem involves calculating the velocity of the image seen in a concave mirror when a metal sphere is moving downwards in a tank with a decreasing water level. The context includes concepts from optics and fluid dynamics, particularly focusing on apparent depth and image formation in mirrors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the real and apparent depths of the sphere and the implications of the water level's change on these depths. Some suggest using matrix methods, while others focus on direct differentiation of depth equations. Questions arise regarding the definition of apparent depth and its measurement.

Discussion Status

Participants are actively exploring different methods to approach the problem, with some providing insights into the definitions and relationships involved. There is a recognition of the need to clarify assumptions about the observer's perspective and the nature of apparent depth. No consensus has been reached yet, but productive lines of inquiry are being pursued.

Contextual Notes

Participants note the complexity introduced by the changing water level and the need to differentiate between various depths and velocities. There is an ongoing discussion about the definitions and implications of apparent depth in the context of the problem.

Saitama
Messages
4,244
Reaction score
93

Homework Statement


Water level in the tank is decreasing at a constant rate of 1 cm/s. A small metal sphere is moving downwards with a constant velocity of 5cm/s. Base of the tank is a concave mirror of radius 40 cm. Find the velocity of the image seen (refractive index of water=4/3)
a)directly
b)after reflection at the mirror

(Answer: a)4 cm/s downwards (b) 59/4 cm/s upwards)

Homework Equations





The Attempt at a Solution


Honestly I don't know where to start with for (a). The water level isn't stationary here. I tried something random but that gives an incorrect result.
Let x be the height of water level above the mirror and y the height of sphere from mirror. The real depth of sphere is (x-y). The apparent depth is ##3/4(x-y)##, differentiating with respect to t, ##3/4(dx/dt-dy/dt)=3/4\times 4=3 cm/s## which is wrong.

Any help is appreciated. Thanks!
 

Attachments

  • velocity of image.png
    velocity of image.png
    26.8 KB · Views: 683
Physics news on Phys.org
I'd probably work it out via the matrixes and put in x-vt for the object position and d-ut for the thickness of the water. The first one just looks like an apparent depth problem.
 
Last edited:
Simon Bridge said:
I'd probably work it out via the matrixes and put in x-vt for the object position and d-ut for the thickness of the water. The first one just looks like an apparent depth problem.

Matrixes? I have never used them in Optics.
 
Pranav-Arora said:
Let x be the height of water level above the mirror and y the height of sphere from mirror. The real depth of sphere is (x-y). The apparent depth is ##3/4(x-y)##, differentiating with respect to t, ##3/4(dx/dt-dy/dt)=3/4\times 4=3 cm/s## which is wrong.

Does "rate of change of apparent depth" represent a velocity with respect to the observer or with respect to the surface of the water?
 
TSny said:
Does "rate of change of apparent depth" represent a velocity with respect to the observer or with respect to the surface of the water?

:confused:
With respect to observer?
 
What is the definition of apparent depth? Would you be able to indicate apparent depth on a diagram?
 
TSny said:
What is the definition of apparent depth? Would you be able to indicate apparent depth on a diagram?

I don't know how to define the apparent depth but I can indicate that on a diagram.
 
Is apparent depth measured from the observer to the image or from the surface of the water to the image?
 
TSny said:
Is apparent depth measured from the observer to the image or from the surface of the water to the image?

From the surface of water.
 
  • #10
Pranav-Arora said:
From the surface of water.
Yes

At any instant of time, how could you write an expression for the distance from the observer to the image in terms of the apparent depth?
 
Last edited:
  • #11
TSny said:
Yes

At any instant of time, how could you write an expression for the distance from the observer to the image in terms of the apparent depth?

Assuming that the height of observer from the surface of water is h, the distance is ##h+3/4(x-y)##. Differentiating this and substituting values gives the correct result.

I will be trying the b) part tomorrow morning. Its getting late here. :)
 
  • #12
Pranav-Arora said:
Matrixes? I have never used them in Optics.
Never mind, you can just use the formulae for apparent depth and for a concave mirror.
Ray Transfer Matrix analysis is often useful for composite systems ... but, the way your problem is worked, it's probably easier to just do it directly.

It's the same as normal, but positions are functions of time.
Could you use the position formula and just differentiate it wrt time?
 
  • #13
Okay, I tried the (b) part but I am getting a wrong result.

The position of image is below the mirror at a distance of 20 cm from mirror. The speed of image is 20 cm/s. If x is the height of water level above mirror and y is the distance of image from the mirror, then apparent depth is 3/4(x+y). Distance of image from observer is h+3/4(x+y). Differentiating w.r.t time, i.e ##dh/dt+3/4(dx/dt+dy/dt)=1+3/4(1-20)=-53/4 cm/s##. :confused:
 
  • #14
Pranav-Arora said:
##dh/dt+3/4(dx/dt+dy/dt)=1+3/4(1-20)=-53/4 cm/s##. :confused:

Is ##x## increasing or decreasing with time?
 
  • Like
Likes   Reactions: 1 person
  • #15
TSny said:
Is ##x## increasing or decreasing with time?

Got it! Thanks! :smile:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
944
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
29
Views
2K
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K