Question #1e8c1

1 Answer
Feb 3, 2017

drawn

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

color(blue)(mu_r/v-mu_i/u=(mu_r-mu_i)/R.........[1])

Where

mu_i->"refractive index of the medium of incident ray"

mu_r->"refractive index of the medium of refracted ray"

u->"object distance"

u->"image distance"

R->"radius of curvature of the curved refracting interface "

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

mu_i=1, " "mu_r=n," "u=-2r," "R=-r," "v=v_1(say)

Inserting these in [1] we get

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

color(green)(=>n/v_1+1/(2r)=-n/r+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 "glass" to "air" then

mu_i=n, " "mu_r=1," "u=-r-(2nr)/(2n-1)=-(r(4n-1))/(2n-1),
R=-2r," "v=v_2(say)

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

color(red)(S_R=PB-abs(v_2)=3r-abs(v_2)=3r-(2r(4n-1))/(3n-1))

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

color(red)(=>S_R=(r(n-1))/(3n-1))

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

From "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

color(red)(S_L=PD-abs(v_3)=r-abs(v_3)=r-(2r)/(n+1)=(r(n-1))/(n+1))