2
Magnification produced by a lens m is given by v/u
41.5 cm to the right of the lens
An equi-convex thin lens (??=1.5) of focal length in air as 30cm is sealed into an opening in one end of tank filled with water ((?=1.33). At the end of the tank opposite the lens is a plane mirror, 80 cm distant from the lens. The position of the image formed by the lens-water-mirror system of a small object outside the tank on the lens axis and 90cm to the left of the lens will be 41.5 cm to the right of the lens
300 m
APERTURE D = F/f where d is the aperture of the objective F is the focal length of the objective f is the f-number (f/) of the objective MAGNIFICATION: BY FIELDS M = Alpha/Theta where M is the magnification Alpha is the apparent field Theta is the true field Apparent Field: the closest separation eye can see is 4', more practically 8-25', 1-2' for good eyes. The Zeta Ursae Majoris double (Mizar/Alcor) is 11.75'; Epsilon Lyrae is 3'. True Field (in ?) = 0.25 * time * cos of the declination (in ') = 15 * time * cos of the declination where time is the time to cross the ocular field in minutes A star therefore moves westward at the following rates: 15? /h (1.25?/5 min) at 0? declination 13? /h (1.08?/5 min) at 30? declination 7.5?/h (0.63?/5 min) at 60? declination. MAGNIFICATION: BY FOCAL LENGTHS M = F/f where M is the magnification F is the focal length of the objective f is the focal length of the ocular At prime focus (ground glass), magnification is 1x for each 25 mm of F MAGNIFICATION: BY DIAMETER AND EXIT PUPIL M = D/d where M is the magnification D is the diameter of the objective d is the exit pupil (5-6 mm is best; 7 mm may not produce a sharp outer image)
3 cm
Magnification produced by a lens m is given by v/u
in doing so, the focal length of the eye lens is effectively decreased
A far-sighted man who has lost his spectacles reads a book by looking through a small hole in the sheet of paper. Then in doing so, the focal length of the eye lens is effectively decreased