Exercise 34.15 Enhanced with Solution The thin glass shell shown in the figure (

Question

Exercise 34.15 Enhanced with Solution The thin glass shell shown in the figure (Figure 1) has a spherical shape with a radius of curvature of 14.0 crn, 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. You may want to review (Da pages 1115-1123 For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Image Figure 1 of 3.30 mm K 15.0 cm Part A Calculate the location of the image of this seed. Submit My Answers Give Up Part B Calculate the height of the image of this seed. Submit My Answers Give Up mmm,

Explanation / Answer

cartesian sign convention is used .

part a:

focal length=-radius of curvature/2

==>f=-7 cm

object distance=u=-15 cm

if image distance is v cm,

then using mirror formula:

(1/v)+(1/u)=1/f

==>(1/v)-(1/15)=-1/7

==>v=-13.125 cm

so image of this seed is at a distance of 13.125 cm .

part b:

height of the image=-(image distance/object distance)*object height

=-(13.125/15)*3.3 mm

=-2.8875 mm

so image is inverted.

magnitude of height of image =2.8875 mm

part c:

if shell is reversed, f=7 cm

then v=4.7727 cm

image distance=4.7727 cm

part d:

image height=-(image distance/object distance)*object height

=(4.7727/15)*3.3 mm

=1.05 mm

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