Score: 1 out of 1 C. The distance between the lenses is L = 15 cm, and the focal

Question

Score: 1 out of 1 C. The distance between the lenses is L = 15 cm, and the focal lengths are obj = 1 cm and feye = 8 cm. The object to be viewed, a cell, is placed a distance of 1.154 cm in front of the objective lens. What is the image distance of the objective lens image? cm D. How far is the objective lens image from the eyepiece lens? E. What is the magnitude of the distance between the eyepiece lens and the image that it forms? Is the image formed by the eyepiece lens to the left or the right of the eyepiece lens? F. What is the magnification due to each individual lens? mob,- meve eye G. The total magnification is the product of the magnification due to each individual lens, and it is the ratio of the final image height over the original object height. If you are viewing a cell with a size of 10 pam, what is the size (absolute value) of the image that you view by looking through the eyepiece? size Presbyopia is the tendency to gradually become far-sighted (hyperopic) as you age. If you have normal vision when you are young, you have a near point of 25 cm. A. If the distance between your eye's lens and retina is 1.65 cm, what is the focal length of your eye's lens when you look at an object at your near point? cm B. As you get older, suppose that the near point of your eye increases to 44 cm. What is the focal length of your eye's lens when you look at an object at your near point now? cm

Explanation / Answer

C. for objective lens,

f1 = 1 cm

object distance, p1 = 1.154 cm

Applying 1/f = 1/p + 1/q

1/1 = 1/1.154 + 1/q1

q1 = 7.49 cm .........Ans

D. distance from eyepiece lens = 15 - 7.49 = 7.51 cm

E. now this image form objective lens will work as image for eyepiece lens,

so p2 = 7.51 cm

f2 = 8 cm

1/8 = 1/7.51 + 1/q2

q2 = - 121.7 cm

image distance = 121.7 cm ......Ans

F. m_obj = - q1 / p1 = - 7.49 / 1.154

= - 6.49 ......Ans

m_eye = - q2 / p2 = 121. / 7.51

= 16.2 ........Ans

G. m = m1 m2 = - 6.49 x 16.2 = - 105.2

and m = hi / ho

- 105.2 = hi / (10 um)

hi = - 1052 um

minus tells that image is inverted,

height of image = 1052 um

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