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I need some help with this particular question:
A concave mirror of focal length f is taken.Few drops of water are sprinkled on it.The objest needle and image needle will coincide ...show more
Best AnswerAsker's Choice
I doubt whether you are expected to do all the detailed calculation in an exam – they may just be looking for physical ‘insight’.
You should know that with no water, you would get:
1/u +1/v = 1/f (where u and v are object and image distances).
u = v if object and image are coincident, giving:
2/u = 1/f
u = 2f
The effect of a thin lens of water will be to reduce this slightly, so 3f/2 (answer b) would be the sensible choice.
A rigorous solution is harder. The solution below is a bit of a 'fudge' but you'll get the idea. There is probably a better/simpler way.
A concave mirror with a radius of curvature R, has a focal length f = R/2 (e.g. 1st link). So R = 2f.
The water will form a lens - flat on the top and radius of curvature R on the lower surface. Taking the refractive index of water as 4/3 (a fairly good approximation and required for this question), the focal length of the water-lens, F, is given approximately by the lensmaker's equation (e.g. see 2nd link):
1/F = (n-1)[1/Rf - 1/Rb]
= (4/3-1)[1/(2f) - 1/infinity] (since a flat surface has an infinite radius of curvature)
= (1/3)[1/(2f)]
= 1/(6f)
F = 6f
Light entering the lens is refracted both on way in and on the way out - so it as if it has been through 2 lenses - or one lens with twice the power of the single water-lens; so the effective focal length of the water lens is halved (since power = 1/focal length and the power is doubled); the effective focal length of the water-lens is 6f/2 = 3f.
The combined focal length, L, of the mirror and water-lens is given by:
1/L = 1/f + 1/F
= 1/f + 1/(3f)
= (4/3)/f
The combination of the mirror and water-lens behaves like a single concave mirror of focal length L. So:
1/u + 1/v = 1/L
If the image and object are coincident, them u = v and the equation becomes
1/u +1/u = 1/L
2/u = (4/3)/f
1/u =(2/3)/f
u = (3/2)f
which is the required answer.
You should know that with no water, you would get:
1/u +1/v = 1/f (where u and v are object and image distances).
u = v if object and image are coincident, giving:
2/u = 1/f
u = 2f
The effect of a thin lens of water will be to reduce this slightly, so 3f/2 (answer b) would be the sensible choice.
A rigorous solution is harder. The solution below is a bit of a 'fudge' but you'll get the idea. There is probably a better/simpler way.
A concave mirror with a radius of curvature R, has a focal length f = R/2 (e.g. 1st link). So R = 2f.
The water will form a lens - flat on the top and radius of curvature R on the lower surface. Taking the refractive index of water as 4/3 (a fairly good approximation and required for this question), the focal length of the water-lens, F, is given approximately by the lensmaker's equation (e.g. see 2nd link):
1/F = (n-1)[1/Rf - 1/Rb]
= (4/3-1)[1/(2f) - 1/infinity] (since a flat surface has an infinite radius of curvature)
= (1/3)[1/(2f)]
= 1/(6f)
F = 6f
Light entering the lens is refracted both on way in and on the way out - so it as if it has been through 2 lenses - or one lens with twice the power of the single water-lens; so the effective focal length of the water lens is halved (since power = 1/focal length and the power is doubled); the effective focal length of the water-lens is 6f/2 = 3f.
The combined focal length, L, of the mirror and water-lens is given by:
1/L = 1/f + 1/F
= 1/f + 1/(3f)
= (4/3)/f
The combination of the mirror and water-lens behaves like a single concave mirror of focal length L. So:
1/u + 1/v = 1/L
If the image and object are coincident, them u = v and the equation becomes
1/u +1/u = 1/L
2/u = (4/3)/f
1/u =(2/3)/f
u = (3/2)f
which is the required answer.
Source:
Asker's Rating & Comment
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Wow I totally did not think about how water could form a lens on the mirror!
Thank you very much:)
Other Answers (1)
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from where does the object and image come into the question??????????
you didnt mention any...please explain the question first
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