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.

Explanation / Answer

A). given , height of seed = Hs = 3.30 mm

object distance u = 15 cm

radius of curvature = 9.5 cm

R= 2F ===> f= R/2 = 9.5 / 2 = 4.75 cm

by mirror formula , 1/f = 1/u + 1/v

==) 1/v = 1/4.75 - 1/15 = 0.1438

==) v =6.95 cm

B). magnification m = v/u = 6.95 / 15 = 0.463

so height of seed becomes,

Hs X m = 3.3mm X 0.463 = 1.53 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/v = 1/f - 1/u = 1/(- 4.75) - 1/15 = - 0.277

==) v = - 3.607 cm.

so computing tthe value of v,

v = 3.607 cm.

D). magnification .= v / u = 3.607 / 15 = 0.2405

height of seed = Hs X m = 3.3 X 0.2405 = 0.7936 mm.

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