magnifying glass
DESCRIPTION
Alat OptikTRANSCRIPT
BY
DZ
PURPOSE LEARNING After learnt optical instruments,
student able to identify the principle application of the magnifying glass,
The students can use the magnifying glass well
MAGNIFYING GLASS Magnifying glasses normally are used to
produce images that are larger than the related objects, but they also can produce images that are smaller than the related objects
A converging lens can be used to ignite paper (a). Light entering parallel to the principal axis converges at the focal point of the lens, concentrating solar energy
The two principal rays show that a convex lens forms a virtual image that is upright and larger compared to the object when the object is located between the lens and the focal point.
Because the principal rays are simply part of a model to help locate an image, they do not have to pass through the picture of the lens in a diagram.
In reality, the image is formed only by the light that passes through the actual lens.
Look at the picture
Magnification of the magnifying glasses Observe with the accommodation eyes
M = 𝑆
𝑛
𝑓 + 1
Observe with non accommodation eyes
M = 𝑆
𝑛
𝑓
Example problem A magnifier with a focal length of 30 cm is used to view a 1-cm-tall object. Use ray tracing to determine the location and size of the image when the magnifier is positioned 10 cm from the object.
THE ANSWER Analyze and Sketch the Problem
• Sketch the situation, locating the object and the lens.
• Draw the two principal rays.
Known: Unknown:
do 10.0 cm di ?
ho 1.0 cm hi ?
f 30.0 cm
a. Use the thin lens to determine d1
1
𝑓 =
1
𝑑1
+ 1
𝑑0
do = 𝑓𝑑
𝑜
𝑑𝑜−𝑓
= (30 𝑐𝑚)(10 𝑐𝑚)
(10 𝑐𝑚) −(30 𝑐𝑚)
= - 15 cm( 15 cm away from the lens on the side opposite the object)
Use the magnification equation and solve for image height. M =
ℎ1
ℎ0
= −𝑑
1
𝑑0
h1 = −𝑑
1.ℎ
0
𝑑𝑜
= −(−15 𝑐𝑚)(2 𝑐𝑚)
10 𝑐𝑚
= 3 cm (3 cm tall)
The positive means that the image is erect.
ConcluSion
The image which is formed by magnifying glass is a virtual, erect and magnified.
Thank you for your attention…