Question

the questions ask for the significant figures to be set to 3 along with the units

1.) An engraver uses a magnifying glass (f = 9.20 cm) to examine some work, as in the drawing. The image he sees is located 28.0 cm from his eye, which is his near point. (a) What is the distance between the work and the magnifying glass? (b) What is the angular magnification of the magnifying glass?

2.) A compound microscope has a barrel whose length is 15.9 cm and an eyepiece whose focal length is 1.7 cm. The viewer has a near point located 25 cm from his eyes. What focal length must the objective have so the angular magnification of the microscope is -320?

3.) The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscope is to provide an angular magnification of -84 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece?

4.) The telescope at Yerkes Observatory in Wisconsin has an objective whose focal length is 19.4 m. Its eyepiece has a focal length of 10.0 cm. (a) What is the angular magnification of the telescope? (b) If the telescope is used to look at a lunar crater (diameter = 6950 m), what is the size of the first image, assuming the surface of the moon is 3.77

Answer 1

1) f = 9.20 cm

v = - 28 cm  since image is vertual

1 / f   = 1/ u + 1 / v

1/ u   =   1/ f + 1/ v = 1 / 9.20  + 1/ 28

u = 6.92 cm

b) Angularmagnification is M = ( 1/ f - 1/v ) N

= ( 1/ 9.2 + 1/ 28 ) 28

= 4.04

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2) M = -(L - f_e)N/(f_of_e). The standard formula for this problem, when the near point is at 25 cm is as follows...

M = -L(25cm) / (f_o * f_e)

320= (15.9)(25)/[(f_o)(1.7)]

= 0.731 cm

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