Question

(1) (A)Where does an object need to be placed relative to a microscope for its 0.250-cm focal length objective to produce a magnification of−100?(Give your answer to at least three decimal places.) Where should the 5.00-cm focal length eyepiece be placed to produce a further fourfold (4.00) magnification?

(B)What magnification is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being viewed? What is the overall magnification if an eyepiece that produces a magnification of 8.50✕ is used?

(C)Suppose a certain person's visual acuity is such that he can see objects clearly that form an image 4.00-µm high on his retina. What is the maximum distance at which he can read the 80.0-cm high letters on the side of an airplane? (Assume a 2.00-cm distance between retina and lens.)

(D)The far point of a myopic administrator is 47.2 cm. (Assume a 2.0 cm distance between retina and lens.) What is the relaxed power of his eyes? (Give your answer to at least one decimal place.) If he has the normal 8.50% ability to accommodate, what is the closest object he can see clearly?

Answer 1

1. a. focal length of objective, fo = 0.25 cm
magnification = -100
hence object distance is u
then
m = v/u
then
-100 = v/u
also
1/v - 1/u = 1/f
1/v - 1/u = 100/0.25
1/v = 400 + 1/u
hence
1/m*u = 400 + 1/u
(1/m - 1)/u = 400
u = -0.002525 m = -0.2525 cm
hence the object has to be placed 2.525 mm from the objective

let the eyepiece be at a distance d from the image formed by the objective
then u = -d
v = ?
fe = 0.05 m
m = 4
then
1/v - 1/u = 1/f
1/v = 20 + 1/u = 1/m*u
20 = (1 - 1/m)/d
if m = -4, d = 62.5mm
if m = 4, 37.5 mm

hence tube length for a real final image is 62.55 mm + 252.5 mm = 315.05mm = 31.505cm
for virtual final image, tube length is 29 cm

c. given
v = 2 cm
and image height is -4 um [ -ve sign for real image]
object height = h, image height i
i/h = -4*10^-4 cm / h (in cm ) = v/u = 2 cm / u (in cm )
hence
h/u = -2*10^-4
for, h = 80 cm
80/u = -2*10^-4
u = - 400000 cm = -4000 m = -4 km
so the person can see 80 cm high sign from a distance of 4 km
  
d. given, v = 2 cm
farpoint = 47.2 cm
hence relaxed power is 1/f where f is focal length when looking at far point in meters
1/2 + 1/47.2 = 1/100*f
f = 0.019186991869 m = 1.91869918699 cm
1/f = 52.1186440677966 D

assuming 8.5% ability to accomodate
then minimum power is 52.11864406(1 + 0.085) = 56.5487288051 D
f = 1.76838634 cm
1/2 - 1/u = 1/f
u = -15.2701391 cm
hence near point is 15.27 cm