One end of a long glass rod n 150 is formed into a convex s

One end of a long glass rod (n = 1.50) is formed into a convex surface with a radius of curvature of magnitude 6.10 cm. An object is located in air along the axis of the rod.

(a) Find the image position corresponding to object distance of 22.5 cm from the convex end of the rod.
cm

(b) Find the image position corresponding to object distance of 11.5 cm from the convex end of the rod.
cm

(c) Find the image position corresponding to object distance of 3.1 cm from the convex end of the rod.
cm

Solution

The expression for imaging surface is convex

n1/o + n2/i = (n2-n1)/R

Where

n1 n2 They are the refractive index of the two media, air and rod

o is the object distance

i is the image distance

R is the radius of curvature of the convex surface

Data

n2 =1.5

R = 6.1 cm

Part a)

o = 22.5 cm

n2/i = (n2-n1)/R – n1/o

n2/i = ( 1.50 – 1)/ 6.10 - 1/22.5

n2/i = 0.0769 - 0.0476 = 0.0293

i = n2/0.0293

i = 1.5/0.0293

i = 40 cm

real image

Part b)

o= 11.5 cm

n2/i = ( 1.50 – 1)/ 6.10 - 1/11.5

n2/i = -0.0100

i = - 1.50/0.0100 = -150 cm

virtual image

Part c)

o = 3.1 cm

n2/i = ( 1.50 – 1)/ 6.10 - 1/ 3.1

i = - 6.10 cm