The eyepiece of a microscope has a focal length of 1.25 cm and the objective len

Question

The eyepiece of a microscope has a focal length of 1.25 cm and the objective lens focal length is 1.36 cm.

(a) If the tube length is 15.0 cm, what is the angular magnification of the microscope?


(b) What objective focal length would be required to triple this magnification?
___cm

-A steep cliff west of Lydia's home reflects a 910-kHz radio signal from a station that is 83 km due east of her home. If there is destructive interference, what is the minimum distance of the cliff from her home? Assume there is a 180

Explanation / Answer

assume for relax observation,

(a).

for objective lens,

1/s + 1/s' = 1/f_ob

1/s + 1/(15 - 1.25) = 1/1.36

s = 1.50928 cm

angular magnification is

M = (s'/s)(25/f_ok)

M = ((15 - 1.25)/1.50928)(25/1.25)

M = 182.20 times




(b).

M = 4 (182.206) = 728.82 times

M = (s'/s)(25/f_ok)

728.82 = ((15 - 1.25)/s)(25/1.25)

s = 0.37732 cm

for objective lens,

1/s + 1/s' = 1/f_ob

1/0.3773 + 1/(15 - 1.25) = 1/f_ob

f_ob = 0.36722 cm

2.

Wavelength of 910kHz = 256.5/10.6 = 24.3 metres.
If the signal reflected in phase from the cliff, minimum would be 1 wavelength to cliff and return for non- destructive interference.
So if it inverts in reflection, that would be it. 1/2 wavelength to cliff, 1/2 back phase shifted, maximum destruction.
So, 1/2 of 24.3m = 12.15m.

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