How Does Refractive Index Affect Position Shift in Optics Problems?

[zorro]

Homework Statement

See this link for the question
http://books.google.com/books?id=bPjHS2e6qDUC&lpg=PT163&dq=A%20circular%20disc%20of%20diameter%20d%20lies%20horizontally%20inside%20a%20metallic%20hemispherical%20bowl%20of%20radius%20'a'&pg=PT163#v=onepage&q=A%20circular%20disc%20of%20diameter%20&f=false

(Q.12)

The Attempt at a Solution

I found out the depth of the disc in terms of the quantities given. The disc gets raised due to the refraction at the surface of liquid. How do I relate the shift in the position of the disc with the refractive index?

 

[rl.bhat]

Let AB be the disc of diameter d = 2l, where l is the radius of the disc.
Let OP be the diameter (2R) of the hemisphere. OAP is the right angled triangle.
BOE is the line of sight over the edge without liquid.
When the hemisphere is filled with a liquid of refractive index μ, disc is visible over the edge,when the ray AO moves along OE after refraction.
Draw a normal at O. Let x be the distance of A from the normal and AO = y..
Identify i and r. From the figure you can see that R = l + x. Or x = R - l
sin(r) = x/y.= y/2R
sin(i) = BO/2R.
Hence μ = sin(i)/sin(r) = BO/y ...(1)
sin(i) = (2l + x)/BO. So BO = (2l + x)/sin(i)
y = x/sin(r)
Substitute these values in eq.(1), You will get
μ = (2l + x)/μx.
μ^2 = (2l+x)/x = (2l + R - l)/(R - l) = (R + l)/(R - l)
By taking componendo and devidendo, you will get the final result.

 

[zorro]

Thank you very much for the solution Sir but JEE is already over