Respuesta :
Answer:
final image is 2.4 cm to the right of diverging lens and final magnification is - 1.2
Explanation:
Consider the converging lens first :
f = focal length of the converging lens = 8.00 cm
p = distance of the coin from the lens = 12.0 cm
q = distance of the image of the coin
Using the lens equation
1/f = 1/p + 1/q
1/8 = 1/12 + 1/q
q = 24 cm
So the converging lens forms the image of the coin 24 cm to its right.
magnification is given as
m = -q/p
m = - 24/12
m = -2
So the converging lens forms the image of the coin twice its original size and inverted.
This image formed by the converging lens behaves as object for the diverging lens.
Consider the diverging lens and the image of the coin formed by converging lens
f = focal length of the diverging lens = - 6.00 cm
p = distance of the image of coin from the lens = 24 - 20 = 4 cm to its right
q = distance of the final image of the coin
Using the lens equation
1/f = 1/p + 1/q
1/(-6) = 1/4 + 1/q
q = - 2.4 cm
So the final image is 2.4 cm to the right of diverging lens
final magnification is given as
m' = m (-q/p)
m' = (-2) (- (- 2.4)/4)
m' = - 1.2
Final image of coins placed left to the converging lens by diverging lens is 2.4 cm right to the diverging lens and magnification of the coin's final image is -1.2.
What is focal length of the lens?
The focal length of the lens is length of the distance between the middle of the lens to the focal point.
It can be find out using the following lens formula as,
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Here, (v)is the distance of the image, (u) is the distance of the object, and (f) is the focal length of the lens.
Given information-
The focal length of the converging lens is 8.00 cm.
The focal length of the diverging lens is 6.00 cm.
The distance between the diverging lens and converging lens is 20 cm.
The distance between the coin and converging lens is 12 cm.
As the focal length of the converging lens is 8.00 cm and the distance between the coin and converging lens is 12 cm. Thus, the distance of the image of coin is,
[tex]\dfrac{1}{v}+\dfrac{1}{24}=\dfrac{1}{8}\\v=24[/tex]
Thus the distance of the image of coin is by converging lens is 24 meter to its right side.
As the magnification of image is the ratio of distance of the object to the distance of the image of the object. Thus,
[tex]m=-\dfrac{24}{12}\\m=-2[/tex]
Thus twice the image of the coin made by converging lens is a object for the diverging lens.
As the focal length of the diverging lens is -6.00 cm and the distance between the image of coin and diverging lens is 4 cm(24-20) to its right. Thus, the distance of the image of coin made by the diverging lens is,
[tex]\dfrac{1}{v}+\dfrac{1}{4}=\dfrac{1}{-6}\\v=-2.4[/tex]
Thus, the final location of the final image is 2.4 cm right to the diverging lens.
Magnification of the final image is,
[tex]m_f=-2\times\dfrac{-(-2.4)}{4}\\m_f=-1.2[/tex]
Thus, magnification of the coin's final image is -1.2.
Learn more about the focal length here;
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