Question

An insect 1.3mm tall is placed 0.9mm beyond the focal point of the objective lens of a compound microscope. The objective lens has a focal length of 14mm , the eyepiece a focal length of 30mm .

Where is the image formed by the objective lens?

Give your answer as the distance from the image to the lens. Express your answer using two significant figures.

How tall is the image mentioned in part A?

Express your answer using two significant figures.

If you want to place the eyepiece so that the image it produces is at infinity, how far should this lens be from the image produced by the objective lens?

Express your answer using two significant figures.

Under the conditions of part C, find the overall magnification of the microscope.

Express your answer using two significant figures.

Answer 1

Its given that,

Object distance = o =0.9mm; actual height of object = 1.3mm

focal length of objective lens = fo = 14 mm and that of eyepiece = fe = 30 mm

we know from lens formula that:

1/f = 1/i +1/o

So i = o x f / (o - f )

i = 0.9 x 14 - (0.9 - 14 ) = -12.6/13.1= -0.96 mm

hence i = - 0.96 mm

(B) How tall is the image

We know that

M = -i/o = h'/h

-(-0.96)/0.9 = h'/1.3

h' = 0.96 x 1.3 / 0.9 = 1.386 = 1.39 mm

Hence h' =1.39 mm

(C) we have here, i1 = infinity and fe = 30 mm

again,

1/fe = 1/i1 + 1/o1

1/30mm = 1/infinity + 1/o2

o2 = 30 mm

Hence lens should be 30mm far from the lens.

(D) The m,agnification given by objective and eyepiece will be

Mo = -1/o = -(-0.96)/0.9 = 1.067

Me = 25/fe = 25/30 = 0.833

M(overall) = Mo x Me = 1.067 x 0.833 = 0.89

Hence M(overall) = 0.89