Question

The left end of a long glass rod 8.40 cm in diameter, with an index of refraction 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.20 cm . An object in the form of an arrow 1.53 mm tall, at right angles to the axis of the rod, is located on the axis 25.0 cm to the left of the vertex of the convex surface.

Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.

Find the height of the image of the arrow formed by paraxial rays incident on the convex surface.

Is the image erect or inverted?

Answer 1

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y = 1.53 mm → 0.153 cm

s = 25.0 cm

R = 4.20 cm

n_G = n_b = 1.60

n_Air = n_a = 1

(n_a / s) + (n_b/s ) = (n_b - n_a) / R

Find s’

(1/25) + (1.60/s ) = (1.60 - 1)/4.20

s’ = 15.56 cm

Find y

m = y /y = (-n_a*s ) / (n_b*s)

y = [(-n_a*s’ ) / (n_b*s)] * y

y = [(-1*15.56) / (1.6*25)] * 0.153 = -0.059517 cm →-0.59517mm [omit the – sign if only magnitude is needed]

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