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Wave Optics

Question
CBSEENPH12039046

(a) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision.

(b) The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focused on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece. 

Solution

a) The fig. below shows the formation of image by a compound microscope at least distance of distinct vision.

where, AB = object, A'B' = image formed by objective and A''B'' = image formed by eyepiece.


fo = focal length of objective,

uo = object distance from objective

vo = image distance from objective

D = distance of least distinct vision

L
 = length of the microscope

(b) Total magnification for least distance of clear vision is given by, 
                      Error converting from MathML to accessible text.                       ... (1) 

where,

L is the separation between the eyepiece and the objective


fo is the focal length of the objective,

fe is the focal length of the eyepiece,

D is the least distance for clear vision,

me is the magnification of the eyepiece,

mo is the magnification of the objective. 

Here, m =20, L =14 cm, D = 20 cm, me = 5
Magnification for the eyepiece:
space space space space space space straight m subscript straight e space equals space 5 space equals space straight m subscript straight e space equals space 1 plus straight D over straight f subscript straight e
rightwards double arrow space 5 straight space equals straight space 1 plus 20 over straight f subscript straight e

rightwards double arrow straight f subscript straight e straight space equals straight space 5 straight space cm comma space is space the space focal space length space of space the space eye minus piece.

Now, putting the value of mo and me in equation (1), we get
straight M straight space equals straight space straight m subscript straight o straight m subscript straight e
space rightwards double arrow space 20 space equals space m subscript o 5

rightwards double arrow space m subscript o space equals space 20 over 5 equals space 4

Now comma space using space the space equation comma space

space space space space space space space space space space space space space space m subscript o equals L over f subscript o
We have,
straight f subscript straight o straight space equals straight space 14 over 4 equals 3.5 straight space cm ;is the required focal length of the objective lens.

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