Calculating Velocity of Reflected Image in Concave Mirror

In summary, the 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. The image seen from the mirror is decreasing in apparent depth at a rate of 53/4 cm/s.
  • #1
Saitama
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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!
 

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  • #2
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.
 
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  • #3
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.
 
  • #4
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?
 
  • #5
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?
 
  • #6
What is the definition of apparent depth? Would you be able to indicate apparent depth on a diagram?
 
  • #7
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.
 
  • #8
Is apparent depth measured from the observer to the image or from the surface of the water to the image?
 
  • #9
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?
 
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  • #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?
 
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  • #15
TSny said:
Is ##x## increasing or decreasing with time?

Got it! Thanks! :smile:
 

1. What is the velocity of an image in optics?

The velocity of an image in optics refers to the speed at which an image appears to move across a screen or surface. It is dependent on the distance between the object and the screen, as well as the refractive properties of the medium through which the light is passing.

2. How is the velocity of an image affected by the distance between the object and the screen?

The velocity of an image is directly proportional to the distance between the object and the screen. This means that as the distance increases, the velocity of the image also increases. In other words, the farther an object is from the screen, the faster its image appears to move.

3. Does the medium through which light passes affect the velocity of an image?

Yes, the refractive properties of the medium through which light passes can affect the velocity of an image. For example, if light travels through a denser material, such as water, the velocity of the image will be slower compared to when it travels through air.

4. How is the velocity of an image related to the speed of light?

The velocity of an image is indirectly related to the speed of light. This means that the velocity of an image is affected by the speed of light, but it is not equal to the speed of light. The velocity of an image is dependent on the distance between the object and the screen, whereas the speed of light is a constant value in a vacuum.

5. Can the velocity of an image exceed the speed of light?

No, the velocity of an image cannot exceed the speed of light. According to the theory of relativity, the speed of light is the maximum speed at which any object can travel. Therefore, the velocity of an image, being a property of light, cannot exceed the speed of light.

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