Optics Homework Help: Find x for Image on Self

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To find the value of x for an object placed above water in a concave mirror, it is essential to consider that the object must be at the center of curvature for the image to form on itself. The equation derived indicates that the object distance is influenced by the refractive index of water, leading to the formula x = R - Mu*h. However, the provided book answer x = (R - h) / Mu does not align with the calculations presented. The discussion highlights confusion regarding the combined focal length of the system, which includes the mirror and the water lens, suggesting a need for further clarification on this concept. Ultimately, the conclusion drawn is that the book's answer may be incorrect, proposing an alternative solution of x = (R/μ) - h.
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Homework Statement


Water is poured into a concave mirror of radius of curvature R up to a height h. An object is placed along the principal axis at a distance x above the level of the water. What should be the value of x so that the image of the object is formed on itself?


Homework Equations


For an image to be formed on itself the object must be placed at the centre of curvature of the concave lens.
Since water has been poured into the mirror, the optical length in the medium (water) is Mu*h (where Mu = refractive index of water).


The Attempt at a Solution



Object distance from the pole of the mirror = x + Mu*h
Or R = x + Mu*h
Or x = R - Mu*h

Since the answer given in the book is x = (R - h) / Mu, I am unable to figure out as to where I have gone wrong. Would request for help. Thanks.
 
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Focal length of the water lens is (μ-1)/R.
Light form the object once reflected from the mirror and twice refracted through water lens before returning back to the object.
Combined focal length of the combination is
1/F = 1/Fm + 1/Fw + 1/Fw
 


I could not quite follow the part regarding the combined focal length. Here we have only one lens (water) and one mirror. I would request for the explanation in a little more detail. Thanks.
 


kihr said:
I could not quite follow the part regarding the combined focal length. Here we have only one lens (water) and one mirror. I would request for the explanation in a little more detail. Thanks.
When the rays from the object reflect back to its position they refract twice and reflect once. So the system is a combination of two lenses and one concave mirror. Its combined focal length is in my post.
 


By your method I don't get the answer given in the book. I guess either the answer given in the book is incorrect, or the approach to the solution may require a re-look.
 


(x + h) can be taken as 2F of the system.
Then
2/(x + h) = 2/R + (μ-1)/R + (μ-1)/R .
Solve for x.
The answer in the book appears to be wrong.
It should be
x = (R/μ) - h
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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