Question #1e8c1
1 Answer
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
When viewed from right the refraction will occur in two curved interface.
1) From
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
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
When viewed from left the refraction will occur in one curved interface.
From
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