Consider a spherically convex refracting surface having a radius of curvature of 100cm. An object is placed...

is placed in front of this surface producing a real image that is 80 cm from the vertex of the surface, assuming that both the object and refracting surface are surrounded by air having an index of refraction around 1. When both the object and the refracting surface is immersed in water, determine the distance from the new image to the vertex of the refracting surface and determine if the new image is real or virtual. You may assume that the index of refraction of water is 4/3.





a) 70cm and virtual


b) 70 cm and real


c) 75 cm and virtual


d) 75 cm and real


e) 80 cm and real


f) none of the above

Consider a spherically convex refracting surface having a radius of curvature of 100cm. An object is placed...
I take this to be a thin lens with one spherical surface and one flat surface.


The formula for the focal length is 1/f = (n-1)(1/r).


1/f=1/i+1/o.


Take n for glass n=3/2.


1/i+1/o= (n-1)(1/r)


1/80+1/ = (3/2-1)(1/100)


1/o=1/200-1/80


o=-(133+1/3)





For the lens in the water 1/f = (3/2-4/3)(1/100)


f=600 cm


1/f=1/i+1/o


1/600=1/i+1/[-(133+1/3)]


1/i=1/600+1/(133+1/3)


i = 171 cm





f) None of the above.


There will be a virtual image at 170 cm