IIT JEE , AIEEE Forum - Physics , Optics - doubt in reflection from concave mirror............

bumba

A concave mirror of radius R is kept on a horizontal table.Water of refractive index $ is poured into it upto a height of h.Where should an object be placed so that image is formed on itself?????????

bumba

come on!!!!!!!!!anyone please help!!!!!!!!!

mohnish_khiani

The question seems easy buts its very complex....i have tried it.plz teel me if its correct.

Let the distance of the object from the water = x
The radius of curvature of the lens is = R
The height of the water filled = H

Let us solve this by the method of interfaces

Interface 1: light ray enters water from air :

image dist/obj dist = (mew)2/(mew)1

u = -x
v = ?
(mew)2 = $
(mew)1 = 1
So by the eqation we get Image dist(v) = -$x

Interface 2: Light gets reflected from the concave mirror

v = ?
u = -($x+H)
f= -R/2
Applying mirror formula: 1/v+1/u=1/f

-1/($x+h) +1/v = -2/R

By solving this we get: v = -[($x+H)R / 2$x+2H-R]

Interface 3: Light now goes back to air from water

u = -[($x+H)R / 2$x + 2H - R] - H
u = -[$xR + 2$xH + 2H^2] / 2$x+ 2H - R

v=?
(mew)1=$
(mew)2=1

By applying the formula : image dist/obj dist =(mew)2/(mew)1
We get:

v = -[$xR+2$xH +2H^2] / 2$^2x+ 2$H - R$

Since this image is formed on the object itself the dist of this image and the object from the water is same.So

v = -[$xR+2$xH +2H^2] / 2$^2x+ 2$H - R$ = -x

By solving this eqation we get a quadratic in "x" :

(2$^2)x^2 - (2R$)x+ 2H^2 = 0

By solving this quadratic equation we get:

x = R (+/-) (R^2 + 4H^2) ^ 1/2 / 2$

Actully reading this is very confusing so write it down on a paper neatly to understand better.

pkg1960

Monish,

i think the solution u posted is slightly inconsistent. first of all the answer shud be independant of the height of the water in the concave mirror. i have a shorter way of soving all such problems.

what we do first is, treat the water in the concave mirror as a plano convex lens, the focal lenght of which can be found using the lens makers formula. now we find the effective focal length of the system, by adding the powers using the formula analogus to addition of resistances in parallel.

we know that the image coincides with the oblect wen it is ket at the center of curvature. thus in this case, the ans shud be twice the resultant focal length

bumba

pkg1960,
can we imagine the top surface to be plane and take R=infinity and apply the formula for refraction at a surface.........
moreover,the answer is R-h/$

pkg1960

yes u hav to cosider it to be a plane surface. this dist is measured from which point???

bumba

from the water surface.....

pkg1960

okay i thought from the center of the lens that why i said that it wud be independant of h. but have u tried my method, i think it shud work.

mohnish_khiani

THANKS pkg1960.i'll try it again, but can u plz tel me the mistake in my solution.

pkg1960

i think one flaw in your solution is that u have used the equation for apparent height/depth (the mew1/mew2 equation) in your sol. here the water in the concave mirror will act as a plano convex lens, and not as a slab hence the lens formula shud be applied after u have calculated the effective focal length using the lens makers formula treating R1 as infinity. thus the height shud not come into play here at all. but bumba said that the distance has to be computed from the water surface thus, using my method whatever distance we get, will be the distance from the pole of the mirror. u can accordingly subtract from it the height and thus get the distance from the surface of the water.

bumba

soory pkg but i tried but failed by u r method.can u please solve it for me!!!!!!!!
i am not sure but i have solved it as follows::::
let the distance of the object be u from the water surface.
considering the water surface to be the first part of the lense(as proposed by you)and applying principle of refraction,we get the that the first refracted rat when extended should form the first image at centre of curvature,else the reflected ray fromt the mirror won't be refracted back in the same path as::::
sin i/sinr=$
and if this r=i'(angle of incidence on the plane surface after reflection from the mirror,then only angle of emergence will be equal to i and image and object coincide.
thus,applying the formula for refraction at the first surface,we get
-$/R-h +1/u=$-1/infinity
=>u=R-h/$,which is the required answer.....
please rate me if my process is correct and am waiting for the solution from u r side by u r process and promising to rate u.........

abhishek_khtr

Hey listen man its a damn easy ques see image of an object can be formed on its place when it is at 2F OR C. As medium doesnt affect d F og a mirror but it does of a lens so the damn sure ans is 2F OR C

PLEASE PLEASE RATE ME I WILL BE THANKFUL 2 U

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