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(a) Draw a ray diagram for final image formed at distance of distinct vision (D) by a compound microscope and write expression for its magnifying power. (b)An angular magnification (magnifying power) of 30x is desired for a compound microscope using as objective of focal length 1.25cm and eye piece of focal length 5cm. How will you set up the compound microscope? |
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Answer» Solution :Ray Diagram: (with proper labeling) (a)  Magnifying power m =`(V_(0))/(u_(0)) (1+(D)/(fe))` `m = (L)/(FO)(1+(D)/(fe))` `because m = m_(o)m_(e) = -30 ("VIRTUAL, inverted")` (b) `because f_(o) = 1.25"CM"` `f_(e) = 5.0"cm"` Let us setup a compound microscope such that the final image be formed at D, then `m_(e) = 1+(D)/(fe) = 1 + (25)/(5) = 6` and position of object for this image formation can be calculated - `(1)/(Ve) - (1)/(ue) = (1)/(fe)` `(1)/(-25) - (1)/(ue) = (1)/(5)` `-(1)/(ue) =(1)/(5) + (1)/(25) = (6)/(25)` `ue = (-25)/(6) = -4.17 "cm"` `because m = m_(o) xx m_(e)` `therefore m_(o) = (+Vo)/(uo) = (-30)/(6) = -5` `therefore V = -5u_(o)` `(1)/(Ve)-(1)/(uo) = (1)/(fo)` `(1)/(-5uo)-(1)/(uo) = (1)/(1.25)` `(-6)/(5uo)= (1)/(1.25)` `"uo" = -1.5"cm" rArr "Vo" = 7.5"cm"` `"Tube length"= V_(o) + |u_(o)| = 7.5"cm" + 4.17"cm"` L = 11.67 cm Object be placed at 1.5cm distance from the objective lens. |
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