A compound microscope has lenses of focal length 1cm and 3cm. An object is placed 1.2 cm from the object lens; if a virtual image is formed 25cm from the eye, calculate the separation of the lenses and the magnification of the instrument.
Answer
Imege distance can be calculated as
"v=\\frac{uf_0}{u-f_0}=\\frac{(1))( 1.2) }{1.2-1}\\\\v=6cm"
Object distance
"u^{'}=\\frac{v^{'}f_e}{v^{'}-f_e}=\\frac{(-25) (3) }{-25-3}=2.7cm"
Separation of lense=6+2.7=8.7cm
Magnification
"M=(\\frac{D}{f_e}-1) \\frac{v}{u}=(\\frac{-25}{3}-1) \\frac{6}{1.2}=-46.7"
Comments
Where did the 12 come from while solving for magnification
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