# Question 1e8c1

Feb 3, 2017

The situation as described in the question has been shown in above figure in which the black spot to be viewed is at the point P

Here we will use the following formula for refraction at curve surface to calculate the shift.

FORMULA

$\textcolor{b l u e}{{\mu}_{r} / v - {\mu}_{i} / u = \frac{{\mu}_{r} - {\mu}_{i}}{R} \ldots \ldots \ldots \left[1\right]}$

Where

${\mu}_{i} \to \text{refractive index of the medium of incident ray}$

${\mu}_{r} \to \text{refractive index of the medium of refracted ray}$

$u \to \text{object distance}$

$u \to \text{image distance}$

$R \to \text{radius of curvature of the curved refracting interface }$

When viewed from right the refraction will occur in two curved interface.
1) From $\text{air" to "glass}$ then

${\mu}_{i} = 1 , \text{ "mu_r=n," "u=-2r," "R=-r," } v = {v}_{1} \left(s a y\right)$

Inserting these in [1] we get

$\textcolor{g r e e n}{\frac{n}{v} _ 1 - \frac{1}{- 2 r} = \frac{n - 1}{-} r}$

$\textcolor{g r e e n}{\implies \frac{n}{v} _ 1 + \frac{1}{2 r} = - \frac{n}{r} + \frac{1}{r}}$

color(green)(=>n/v_1=-n/r+1/(2r)=-(2n-1)/(2r)

color(green)(=>v_1=-(2nr)/(2n-1)" where " n>1

2) From $\text{glass" to "air}$ then

${\mu}_{i} = n , \text{ "mu_r=1," } u = - r - \frac{2 n r}{2 n - 1} = - \frac{r \left(4 n - 1\right)}{2 n - 1} ,$
$R = - 2 r , \text{ } v = {v}_{2} \left(s a y\right)$

Inserting these in [1] we get

color(green)(1/v_2-n/(-(r(4n-1))/(2n-1))=(1-n)/(-2r)

color(green)(=>1/v_2+(n(2n-1))/(r(4n-1))=(n-1)/(2r)

color(green)(=>1/v_2=(n-1)/(2r)-(n(2n-1))/(r(4n-1))

color(green)(=>1/v_2=(4n^2-5n+1-4n^2+2n)/(2r(4n-1))

color(green)(=>1/v_2=-(3n-1)/(2r(4n-1))

color(green)(=>v_2=-(2r(4n-1))/(3n-1)

So finally Shift of point P when viewed from right will be

$\textcolor{red}{{S}_{R} = P B - \left\mid {v}_{2} \right\mid = 3 r - \left\mid {v}_{2} \right\mid = 3 r - \frac{2 r \left(4 n - 1\right)}{3 n - 1}}$

color(red)(=>S_R=r/(3n-1)(9n-3-8n+2)

$\textcolor{red}{\implies {S}_{R} = \frac{r \left(n - 1\right)}{3 n - 1}}$

When viewed from left the refraction will occur in one curved interface.

From $\text{glass" to "air only}$ then

mu_i=n, " "mu_r=1," "u=-r," "R=-2r," "v=v_3say)

Inserting these in [1] we get

color(blue)(1/v_3-n/-r=(1-n)/(-2r)

color(blue)(=>1/v_3+n/r=-1/(2r)+n/(2r)

color(blue)(=>1/v_3=-1/(2r)+n/(2r)-n/r

color(blue)(=>1/v_3=-1/(2r)(1-n+2n)

color(blue)(=>v_3=-(2r)/(n+1)#

So finally Shift of point P when viewed from left will be

$\textcolor{red}{{S}_{L} = P D - \left\mid {v}_{3} \right\mid = r - \left\mid {v}_{3} \right\mid = r - \frac{2 r}{n + 1} = \frac{r \left(n - 1\right)}{n + 1}}$