A symmetric biconvex lens of radius of curvature and made of glass of refractive index 1.5 is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real inverted image coincides with the needle itself.The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of X and Y
Answers
And it can come normal to plane mirror if it comes from infinity i.e. it was placed at focus...
So x and y are considered as focuses of repectives..
Hey !!
When a liquid is placed on top of the plane mirror and convex lens over it, then this whole system would become a combination of convex lens of glass and planoconcave lens of liquid. This is shown in the figure.
Let focal length of convex lens = f₁
Focal length of planoconcave liquid lens = f₂
combined focal length = F
In both cases image coincides with needle, hence ray is normal to plane mirror. So, needle position is focal lengths of convex lens and combined system respectively.
Now, According to the question,
f₁ = y unit
F = x unit
We also know that for combination of two lenses
1/F = 1/f₁ + 1/f₂
= 1/f₁ = 1/F - 1/f₁
= 1/f₂ = 1/x - 1/y
f₂ = xy / y - x
For glass lens, let R₁ = R, R₂ = -R. From lens maker formula
1/f = (n-1) (1/R - (1/R))
= 1/y = (1.5 - 1) (1/R + 1/R)
= 1/y = 1/R
R = y unit
For liquid planoconcave lens
R₁ = -R , R₂ = infinity
For lens maker formula
1/f₂ = (n₁ - 1) ( -1/R - 1/infinity)
= 1 - n₁ = y - x / x
n₁ = 1 - y - x / x
n₁ = x - y + z / x
n₁ = 2x - y / x
GOOD LUCK !!
