Calculating Velocity of Reflected Image in Concave Mirror

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The discussion revolves around calculating the velocity of the image seen in a concave mirror when a metal sphere is submerged in water that is decreasing at a constant rate. The participants explore the apparent depth of the sphere and how it changes with time, leading to confusion about the correct application of formulas. They clarify that the apparent depth is measured from the water's surface and discuss how to express the distance from the observer to the image. Ultimately, they derive the velocities for both direct observation and reflection, arriving at the correct results for the problem. The conversation highlights the importance of understanding the concepts of apparent depth and differentiation in optics.
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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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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?
 
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  • #15
TSny said:
Is ##x## increasing or decreasing with time?

Got it! Thanks! :smile:
 

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