Question

A glass rod 96 cm long with an index of refraction of 1.6 has its ends ground to convex spherical surfaces of radii 8 cm and 16 cm. A point object is in air on the axis of the rod 19.1 cm from the end with the 8 cm radius.

(a) Find the image distance due to refraction at the first surface. (Include the sign.)
___cm

(b) Find the location of the final image distance due to refraction at both surfaces.
___cm (from the 8 cm diameter surface)
(c) Describe the final image.

real

virtual

inside the rod

outside the rod

Answer 1

given that

n₁ = 1 , n₂ = 1.6

r = 8 cm , s = 19.1 cm

the image distance due to first surface is

s' = n₂*r*s/[(n₂ - n₁)s - n₁r] = (1.6)(8)(19.1)/ [(0.6)(19.1)-8] = 70.66 cm

the object for the second surface is the image formed by the first surface,

thus, we have

n₁ = 1.6 , n₂ = 1

r = -16 cm , s = 32 cm

thus the image distance due to refraction at both surfaces is

s' = n₂*r*s/[(n₂ - n₁)s - n₁r] = (1)(-16)(32)/ [(-0.6)(32)-(1.6)(-16)] = -80 cm

the final image formed is virtual