Herewith is the question: A farsighted person can read printing as close as 25.0
ID: 1757391 • Letter: H
Question
Herewith is the question:A farsighted person can read printing as close as 25.0cmwhen she wears contacts that have a focal length of 45.4cm. On day,however, she forgets her contacts and uses a magnifying glass. Ithas a maximum angular magnification of 7.50 for a young person witha normal near point of 25.0cm. What is the maximum angularmagnification that the magnifying glass can provide forher? I guess the formula we need the angular magnification of amagnifying glass: M=(1/f - 1/di)N where M is the angular f is the focal lengh di is the virtual image distance N is the Near Point Now where? Thanks! Herewith is the question:
A farsighted person can read printing as close as 25.0cmwhen she wears contacts that have a focal length of 45.4cm. On day,however, she forgets her contacts and uses a magnifying glass. Ithas a maximum angular magnification of 7.50 for a young person witha normal near point of 25.0cm. What is the maximum angularmagnification that the magnifying glass can provide forher? I guess the formula we need the angular magnification of amagnifying glass: M=(1/f - 1/di)N where M is the angular f is the focal lengh di is the virtual image distance N is the Near Point Now where? Thanks! I guess the formula we need the angular magnification of amagnifying glass: M=(1/f - 1/di)N where M is the angular f is the focal lengh di is the virtual image distance N is the Near Point Now where? Thanks!
Explanation / Answer
I guess the formula we need the angular magnification of amagnifying glass: M=(1/f - 1/di)N where M is the angular f is the focal lengh di is the virtual image distance N is the Near Point Now where? Thanks!the focal length of magnifying glass=f
now 7.5=D/f
f= D/7.5=25/7.5=3.3333cm
u=25cm
so 1/v=1/f+1/u=1/3.333+1/25=>v=2.9409cm
now substituting the values in the above expression we getM=(1/3.333-1/2.9409)25
so M=(0.3+0.3400)25=16