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

Question

24.17 The thin glass shell shown in the figure (Figure 1) has a spherical shape with a radius of curvature of 10.0 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.

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

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

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

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

Explanation / Answer

Hs = 3.30mm

object distance = 15 cm. = u

radius of curvature = 10.0 cm .

R = 2F

f = R / 2 = 5.0 cm.

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A)

1/f = 1/u + 1/v

since the mirror is concave, the focal length is taken positive.

1 / 5.0 = 1 / 15 + 1/v

v = 7.5 cm

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B)

magnification = v / u

= 7.5 / 15 = 0.5

so height of seed becomes,

Hs * m = 3.3mm X 0.5

= 1.65 mm

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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 / -5.0 = 1 / 15 + 1/v

v = 7.5 cm

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D)

magnification .= v/u = -3.75/ 15 = 0.25

height of seed = Hs * m = 3.3 X 0.25 = 0.825 mm.

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