Find the focal length of the magnifying glass?

Question

if you can see the butterfly clearly when your eye is 5.5 cm away from the glass and the glass is 25 cm away from the butterfly. The formula is 1/s + 1/t ≈ 1/f.


a) focal length is about 30.5 cm
b) focal length is about 4.5 cm
c) focal length is about 4.2 cm


which option is a true answer ?

Answer 1

Your formula 1/s + 1/t = 1/f is correct in its form, although I haven't seen it with those particular variables; the object and image lengths are usually called R1 and R2 or s1 and s2, or sometimes s and s'. So I'm not sure which of s and t is supposed to be the object's location and which is the observer's position, but luckily it doesn't matter.

1/5.5 + 1/25 = 1/f ==> 0.18 + 0.04 = 1/f ==> 1/f = 0.22 ==> f = 1/0.22 = 4.51

The focal length is 4.51 cm, so choice b) about 4.5 cm is the correct answer.

Answer 2

Converging (i.e., convex or magnifying) lenses act to focus parallel rays of light that are intercepted by their area onto a point (or line, in the case of cylindrical lenses) lying in the focal plane of the lens.

See http://hypertextbook.com/physics/waves/lenses/ for a
good explanation. To increase the amount of energy focused at the focal point, one needs to increase the amount of sunlight intercepted by the lens. Obviously, the easiest way to do this is to increase the size (aperture) of the lens. The "light gathering" power of a lens is directly related to the area of the lens. This is why larger telescopes can "see" fainter objects.

Increasing the curvature of the lens changes the "focal length" of the lens (the distance between the focal plane and the center of the lens); it does *not* increase the amount of energy focused at the focal point. Higher curvature will decrease the focal length, bringing the "hotspot" closer to the lens. Generally, this is not want you want for your application. Instead, you would probably prefer to have the "hotspot" located some conveniently large distance
from the lens itself because this makes it easier to protect the lens, as well as to manipulate the object being heated.

There are about 1000 watts/(square meter) of sunlight at high noon on a clear day at the surface of the Earth. This is roughly the same power as a heating element on an electric stove. A circular lens with a diameter of 7 cm (about 2.75 inches) has an area of about 38.5 cm^2, or 0.00385 m^2. Assuming a solar insolation of 1000W/m^2, this lens
would concentrate 0.00345*1000 = 3.45 Watts onto the focal point. A lens with 10 times that diameter has an area 100 times larger, and would intercept and concentrate 100 times more energy (345 Watts) onto the focal point.

Of course, real lenses don't actually focus onto a true "point";
optical abberations and diffraction place lower limits on the size of the "hotspot". In addition, sunlight is made up of all colors of light. All real lenses suffer some degree of chromatic abberation (resulting from the fact that different colors/wavelengths of light are refracted different amounts by the lens), so that different colors are focused onto slightly different focal planes..

You may find the following links of interest:

http://sci-toys.com/scitoys/scitoys/light/marshmallows/solar_roaster.html

http://answers.google.com/answers/threadview?id=525247

Answer 3

1/s + 1/t = 1/f
1/25 + 1/5.5 = 1/f
(25 +5.5)/ 25 × 5.5 = 1/f
30.5/25 × 5.5 = 1/f
6.1/27.5 = 1/f
Threfore f = 27.5/6.1
= 4.5 cm Option (b)

Answer 4

b