1.An object 2.5cm high is placed 10cm from a converging lens. if the local length of the lens is 5cm,
a. How far from the lens will the image be formed?
b. How high is the image formed?
c. What is the total magnificatio of the object?
2. A terrestrial telescope is used to view an object 1.6km away. Calculate the magnification power of the telescope given that the focal length of its eyepiece is 5.08cm and of its objective len, 12.7cm
3. An object is located 3cm from a convex mirror and produces a virtual image at a distance of 5cm.
a. what is the focal length of the mirror?
b. if the object is 1.6cm high, what is the size of the image?
What effort have you made on these problems? What is the difficulty applying the standard lens equiation?
In 1, do not confuse the words focal and local.
im not sure
1. To solve this problem, we can use the lens equation:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens
Given:
f = 5cm
u = -10cm (negative sign because the object is placed on the opposite side of the lens)
a. How far from the lens will the image be formed?
To find v, we can rearrange the lens equation:
1/v = 1/f - 1/u
Substituting the given values:
1/v = 1/5 - 1/(-10)
Simplifying:
1/v = 1/5 + 1/10
1/v = 3/10
Taking the reciprocal of both sides:
v = 10/3
Therefore, the image will be formed 10/3 cm from the lens.
b. How high is the image formed?
To find the height of the image, we can use the magnification formula:
magnification = height of the image / height of the object = -v / u
Substituting the given values:
magnification = -v / u
magnification = -(10/3) / (-10)
magnification = 1/3
The height of the image is one-third the height of the object.
c. What is the total magnification of the object?
To find the total magnification, we multiply the magnification of the lens and the magnification of the object:
total magnification = lens magnification * object magnification
total magnification = -(10/3)/(-10) * (-10/2.5)
total magnification = 4
The total magnification of the object is 4.
2. To calculate the magnification power of the telescope, we can use the formula:
Magnification Power = focal length of objective lens / focal length of eyepiece
Given:
Focal length of objective lens = 12.7cm
Focal length of eyepiece = 5.08cm
Magnification Power = 12.7 / 5.08
Magnification Power ≈ 2.5
The magnification power of the telescope is approximately 2.5.
3.
a. To find the focal length of the convex mirror, we can use the mirror equation:
1/f = 1/v - 1/u
Where:
f = focal length of the mirror
v = image distance from the mirror (negative for virtual image)
u = object distance from the mirror
Given:
u = 3cm
v = -5cm
Substituting the given values into the mirror equation:
1/f = 1/(-5) - 1/3
Simplifying:
1/f = -1/5 - 1/3
1/f = -8/15
Taking the reciprocal of both sides:
f = -15/8
Therefore, the focal length of the convex mirror is -15/8 cm.
b. To find the size of the image, we can use the magnification formula:
magnification = height of image / height of object = -v / u
Given:
u = 3cm
v = -5cm
Height of object = 1.6cm
Substituting the given values into the magnification formula:
magnification = -v / u
magnification = -(-5) / 3
magnification = 5/3
To find the size of the image, we multiply the magnification by the height of the object:
Size of image = magnification * height of object
Size of image = (5/3) * 1.6
Size of image ≈ 2.67cm
Therefore, the size of the image formed by the convex mirror is approximately 2.67cm.