Question #197302

A microscope has an objective of focal length 1.0cm and an eyepiece of focal length 8.0cm.Both lenses are separated by a distance of 12.0 cm. A tiny object 1mm long is placed 1.2cm from the objective. Find the position and size of the final image


1
Expert's answer
2021-05-24T14:11:41-0400

u=u = distance of object from objective lens.

v=v = image formed by objective lens.

f0=f_0 = objective focal length.

fe=f_e = eye piece focal length.

We know,


1f0=1u+1v\dfrac{1}{f_0} = \dfrac{1}{u}+\dfrac{1}{v}


1=1v11.21 = \dfrac{1}{v}-\dfrac{1}{1.2}


1v=2.21.2\dfrac{1}{v} = \dfrac{2.2}{1.2}


v=1.22.2v = \dfrac{1.2}{2.2}


v=0.54cmv= 0.54cm

Distance of object from eye lens =120.54=11.46cm= 12-0.54 = 11.46cm

Now,

u1=u_1= distance of object from eye piece = -11.46cm

v1=v_1 = distance of image formed by eye piece

Hence,


1fe=1u1+1v1\dfrac{1}{f_e} = \dfrac{1}{u_1}+ \dfrac{1}{v_1}


18=111.46+1v1\dfrac{1}{8} = -\dfrac{1}{11.46} + \dfrac{1}{v_1}


v1=4.76cmv_1 = 4.76cm


We know,

Magnification =hih0=f0fe= -\dfrac{h_i}{h_0} = \dfrac{f_0}{f_e}


hi1=18-\dfrac{h_i}{1} = \dfrac{1}{8}


hi=0.125mmh_i = 0.125mm


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