Question

# The left end of a long glass rod 8.60 cm in diameter, with an index of...

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.

s′

s′

=

nothing

cm

Part B

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

y′

y′

=

nothing

mm

Part C

Is the image erect or inverted?

Is the image erect or inverted?
 erect 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

#### Earn Coins

Coins can be redeemed for fabulous gifts.