An insect 1.2 mm tall is placed 1.0 mm beyond the focal point of the objective l
ID: 1399833 • Letter: A
Question
An insect 1.2 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens has a focal length of 14 mm , the eyepiece a focal length of 20 mm.
PART A: Where is the image formed by the objective lens? Give your answer as the distance from the image to the lens. Express your answer using two significant figures.
PART B:How tall is the image mentioned in part A? Express your answer using two significant figures.
PART C: If you want to place the eyepiece so that the image it produces is at infinity, how far should this lens be from the image produced by the objective lens? Express your answer using two significant figures.
PART D: Under the conditions of part C, find the overall magnification of the microscope. Express your answer using two significant figures.
Explanation / Answer
The compound microscope has two lenses, the objective lens and the eye piece.
(a) the distance of image from objective lens can be calculated by lens formula, that is
1/f=1/p+1/q where f is the focal length, P is the object distance and q is the image distance.
for image formation, we can rearrange the formula. that is 1/q=1/f-1/p or q=pf/p-f
putting the values we get q= 210mm (the image is formed 210 mm away from the objective lens)
(b) the height of image can be found by the magnification formula, that is M0=hi/ho where hi and ho are the height of image and height of object respectively. magnification can also be expressed interms of distances, that is M0=q/p. therefore comparing them yields, q/p=hi/ho. rearranging for hi, we get hi=ho(q/p). putting the values we get hi= 16.8 mm. since p is given and we calculated the q, therefore M0=210mm/15mm = 14 (the image is magnified to 14 times the size of original insect)
(c) for the image produced at infinity by eye piece, the image from objective lens should be at the focal point of eye piece. the focal length of eye piece is given, therefore, the eye piece will be q+fe = 230 mm away from objective lens. (where fe is the focal length of eye piece)
(d) the formul for total magnification is given by Mtotel = M0Me where Me is the magnification of eye piece (Me=25/fe) therfore M= Mo 25/fe. putting the values we get M=17.5