The left end of a long glass rod 8.60 cm in diameter, with an index of refraction 1.58, is ground and polished to a convex hemispherical surface with a radius of 4.30 cm . An object in the form of an arrow 1.54 mm tall, at right angles to the axis of the rod, is located on the axis 23.0 cm to the left of the vertex of the convex surface. 
Part A Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.
SubmitRequest Answer Part B Find the height of the image of the arrow formed by paraxial rays incident on the convex surface.
SubmitRequest Answer Part C Is the image erect or inverted? Is the image erect or inverted?

for spherical refracting surface,Using equation:
na/s + nb/s' = (nb  na)/R
Given that
R = Radius of curvature = 8.60/2 = 4.30 cm
na = refractive index of air = 1.00
nb = refractive index of glass = 1.58
s = object distance = 23.0 cm
s' = image distance = ?
Using these values:
1/23 + 1.58/s' = (1.58  1)/4.3
1.58/s' = 0.58/4.3  1/23
s' = 1.58*4.3*23/(23*0.58  4.3)
s' = 17.3 cm = image distance
Part B.
Now magnification is given by:
M = na*s'/(nb*s)
M = 1*17.3/(1.58*23)
M = 0.476
Now Magnification is also given by:
M = h'/h
h = object height = 1.54 mm
h' = image height = ?
h' = M*h
h' = 0.476*1.54
Image height = 0.73 mm
Part C.
Since M < 0, So image is inverted
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