Question

The compound microscope consists of two converging lenses: the objective and the eyepiece. Suppose the focal length f_1 of the objective is 20 cm, and the focal length f2 of the eyepiece is 20 cm. The separation between the objective and eyepiece is 66 cm. The object is located 30 cm in front of the objective. (These numbers are not realistic for a real microscope.) a. Calculate the location and magnification of the image, and state whether it is a real or virtual, upright or inverted. b. Construct a ray tracing diagram. The distances between optical components must be drawn to scale. c. Normally your eye is located fairly close to the eyepiece. Looking at the position of the image and knowing the near point of a normal eye, comment on how comfortable the scientist would be to look at such an image for long periods of time.

Answer 1

The object distance u, image distance v and focal length of a lens are related by the expression

$\frac{1}{f} = \frac{1}{v}+ \frac{1}{u}$

for the objective, $\frac{1}{f_1} = \frac{1}{v_1}+ \frac{1}{u_1}$

=> $\frac{1}{20} = \frac{1}{v_1}+ \frac{1}{30}$

=> v₁ = 60 cm

and magnification m₁ = -v₁/u₁ = - 60/20 = -3

distance between the objective and eyepiece is 66 cm

therefore, u₂ = 66 - 60 = 6 cm

so, $\frac{1}{20} = \frac{1}{v_2}+ \frac{1}{6}$

=> v₂ = - 8.57 cm

and so the magnification will be: m₂ = -(-8.57)/6 = 1.43

and therefore the total magnification of the system is: m = m₁m₂ = -3(1.43) = - 4.29.

the final image will be virtual, magnified and upright.

The near point of a human eye is about 25 cm which basically means that within 25 cm from the eye, we can bring the image into focus and see a sharp image. Since this image due to the eyepiece is just 8.57 cm from the eye, therefore the scientist will suffer discomfort if viewing for prolonged period of time.