1. What can one say about the image produced by a thin lens that produces a positive magnification? a. It is real…
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Question “1. What can one say about the image produced by a thin lens that produces a positive magnification? a. It is real…”
1. What can one say about the image produced by a thin lens that
produces a positive magnification?
a. It is real and inverted
b. It is real and erect.
c.It is virtual and inverted.
d. It is virtual and erect.
2.If
the diameter of a lens is reduced, what happens to the
magnification produced by the lens?
a. It increases
b. It decreases
c. It is unchanged
Answer
This problem can be solved using three concepts: magnification, image elevation and object height.
Use the conditions of the sign convention for magnification caused by thin lenses.
To prove that magnification is not related to radius of curvature of the lens, you can use the expression magnification in terms image and object height.
A coordinate system is constructed if the direction of light falling on the lens is from left-to-right. X and y are the coordinate axes. This is the sign convention.
The region to the right and left of the origin are considered positive along the x-axis.
The y axis indicates that the region upwards from the origin is considered positive, while the region downwards are considered negative.
Magnification is measured in image height and object height.
m = \frac{{{h_i}}}{{{h_o}}}Here, the magnification of $math_tag_1 is used.
Magnification is measured in terms of object distance and image distance.
m = - \frac{{{d_i}}}{{{d_o}}}The image distance is $math_tag_3
(1)
Magnification can be described as:
m = \frac{{{h_i}}}{{{h_o}}}Magnification is measured in terms of object distance and image distance.
m = - \frac{{{d_i}}}{{{d_o}}}These are the wrong options
It is both real and inverted
Positive magnification means that the image cannot be reversed.
b.It’s real and erect
A positive magnification requires that the image distance be negative. A negative image distance can not produce a real image. This statement is false.
c.It can be virtualized and inverted
This is because the negative magnification can’t produce an inverted image.
The best option is:
d.It’s virtual and erect
Because a positive magnification creates a virtual upright image, the above option is correct.
(2)
Magnification is measured in terms of distance between objects and images.
m = - \frac{{{d_i}}}{{{d_o}}}These are the wrong options
It increases
This is because magnification depends upon image height and object elevation, which are both independent of radius curvature.
b.It drops
Because magnification depends on radius of curvature, the above option is not correct.
The best option is
It is unaltered
Because magnification does not depend on the radius of curvature, the above option is correct.
Part 1: Ans
The thin lens produces a virtual image that is erect.
Conclusion
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