The thin glass shell shown in the figure (Figure 1) has a spherical shape with a

Question

The thin glass shell shown in the figure (Figure 1) has a spherical shape with a radius of curvature of 9.50 cm , and both of its surfaces can act as mirrors. A seed 3.30 mm high is placed 15.0 cm from the center of the mirror along the optic axis, as shown in the figure.

Part A Calculate the location of the image of this seed.

Part B Calculate the height of the image of this seed.

Part C Suppose now that the shell is reversed. Find the location of the seed's image.

Part D Find the height of the seed's image.

PLease type in the answer and not take a picure of a notebook .

Explanation / Answer

Hs = 3.30mm

object distance = 15 cm. = u
radius of curvature = 9.50 cm .
R = 2F

f = R / 2 = 4.75 cm.

A)

1/f = 1/u + 1/v
since the mirror is concave, the focal length is taken positive.

1 / 4.75 = 1 / 15 + 1/v
v = 6.95 cm

B)

magnification = v / u

= 6.95 / 15 = 0.46
so height of seed becomes,

Hs * m = 3.3mm * 0.46

= 1.52 mm

C)

shell is reversed, then the glass piece becomes a convex mirror of the same focal length
so, now since it is a convex mirror, the focal length is taken - ve .

1/f = 1/u + 1/v

1 / -4.75 = 1 / 15 + 1/v

v = -3.60 cm.

D)

magnification .= v/u = -3.60/ 15 = 0.24
height of seed = Hs * m = 3.3 X 0.24 = 0.793 mm.

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