The left end of a long glass rod 8.00 cm in diameter and with an index of refrac

Question

The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00 cm . An object in the form of an arrow 1.20 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. Part B Find the height of the image formed by paraxial rays incident on the convex surface. Part C Is the image erect or inverted?

Explanation / Answer

given that

n1 = 1

n2 = 1.60

u = -23 cm

R = 4 cm

we know that

n2/v - n1/u = (n2-n1) / R

1.60/v - 1/(-23) = (1.60-1) / 4

1.60/v + 0.043 = 0.15

v = s' = 14.95 cm

(b)

we know that

magnification = image height/object height = - image distance / object diatance

h2 / h1 = - v / (-u)

h2 / 1.20 = 14.95 / 23

h2 = 0.78 mm

(c)

because height of image is positive means the image is erect

answer

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