Answer to Question #162620 in Optics for friedel craft

Question #162620

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.


1
Expert's answer
2021-05-27T10:17:10-0400

Answer

Imege distance can be calculated as

v=uf0uf0=(1))(1.2)1.21v=6cmv=\frac{uf_0}{u-f_0}=\frac{(1))( 1.2) }{1.2-1}\\v=6cm



Object distance

u=vfevfe=(25)(3)253=2.7cmu^{'}=\frac{v^{'}f_e}{v^{'}-f_e}=\frac{(-25) (3) }{-25-3}=2.7cm


Separation of lense=6+2.7=8.7cm


Magnification

M=(Dfe1)vu=(2531)61.2=46.7M=(\frac{D}{f_e}-1) \frac{v}{u}=(\frac{-25}{3}-1) \frac{6}{1.2}=-46.7






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Comments

Chioma
26.05.21, 23:07

Where did the 12 come from while solving for magnification

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