Question

This is a two part question.

Part 1. To look closely at a small object, such as an insect or a crystal, you bring it close to your eye, making the subtended angle and the retinal image as large as possible. But your eye cannot focus sharply on objects that are closer than the near point, so the angular size of an object is greatest when it is placed at the near point (25 cm away from a “normal” eye). To enlarge the object even more, one uses a magnifying glass, which is a converging lens that produces a virtual image that is larger and further away from the eye than the object is. The usefulness of the magnifying glass is given by the angular

magnification

m ang= θ′/θ

which is the ratio of the angular size θ′ of the image seen through a magnifier when the object is placed at the magnifier’s focal point to the largest angular size θ of the object (when you view it from the near point)1. Find the angular magnification of a magnifying glass made of a converging lens with a focal length of 5 cm, assuming that the person who uses it has perfect vision.

Part 2. In an astronomical telescope the image formed by an objective is viewed through an eyepiece. To have the final image formed at infinity (for most comfortable viewing by a normal eye), the distance between objective and eyepiece, which is the length of the telescope, is made equal to the sum of the focal lengths of objective f1 and eyepiece f2. Find the angular magnification (see the definition above) of a telescope with f1 = 100 cm and f2 = 10 cm, which is defined (similar to above) as the ratio of the angle subtended at the eye by the final image to the angle subtended at the (unaided) eye by the object: mang = θ′/θ. Find the height of the image formed by the objective of a 20-story building 60 m tall, 3 km away. What is the angular size of the final image as viewed by an eye very close to the eyepiece?

Foot Note 1: One should not confuse the angular magnification mang with the (lateral) magnification m = −s′/s. Angular magnification is the ratio of the angular size of an image to the angular size of the corresponding object; lateral magnification refers to the ratio of the height of an image to the height of the corresponding object. For the situation discussed in the problem, the angular magnification is typically from 2 to 10. The lateral magnification, however, is infinite in this case, because the virtual image produced by the magnifying glass is at infinity (s′ = ∞), but that does not mean that objects looks infinitely large through the magnifier!

Answer 1

Part 1)

Angular magnification $m_{ang}=\frac{\theta'}{\theta}$

Angular size of object ${\theta}=\frac{h_o}{D}$   

where $h_o$ is the object distance from convex lens , $h_o=5\,cm$ since the object is placed at focal point of lens .

Angular size of image ${\theta'}=\frac{h_o}{f}$

Angular magnification    $m_{ang}=\frac{\theta'}{\theta}=\frac{\frac{h_o}{f}}{\frac{h_o}{D}}=\frac{D}{f}$

$m_{ang}=\frac{25}{5}=5$

Part 2)

Object ( building ) is at a distance of $3\,km=3000\,m$ from objective.

That is angular size of object is $\theta_1=\frac{60}{3000}$ ------------(1)

A real image of height is  formed by objective at a the focal point of objective ( 100 cm=1 m) , which also makes the same angle with the center of objective.

so $\theta_1=\frac{h}{f_1}=\frac{h}{1}$ -----------(2)

from (1) and (2) $\frac{h}{1}=\frac{60}{3000}$

${h}=\frac{1}{50}=0.02 \,m =2\,cm$

Height of the image formed by objective is

This image acts as an object to eyepiece.

Angular size of this image with the eyepiece is ( focal length $f_2=10\,cm=0.1\,m$ )

$\theta_2=\frac{h}{f_2}$

$\theta_2=\frac{0.02}{0.1}=0.2\,rad$

the angular size of the final image as viewed by an eye very close to the eyepiece is $\theta_2=0.2\,rad$