Answer :
Answer:
E) true. The image is always virtual and erect
Explanation:
In this exercise we are asked to find the correct statements,
for this we can use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p the distance to the object and q the distance to the image
In diverging lenses, the focal length is negative and the image is virtual and erect
In convergent lenses, the positive focal length, if the object is farther than the focal length, the image is real and inverted, and if the object is at a shorter distance than the focal length, the image is virtual and straight.
With this analysis let's review each statement
A) False. The image is right
B) False. The type of image depends on where the object is with respect to the focal length
C) False. The real image is always inverted
D) False. The image is always virtual
E) true. The image is always virtual and erect
A diverging lens always produces a virtual erect image.
The general lens formula is given as;
[tex]\frac{1}{F} = \frac{1}{U} + \frac{1}{V}[/tex]
Where;
- U = object distance
- V = image distance
- F = focal length of the lens
A lens can be converging or diverging.
A converging lens produces a virtual image when the object is placed in front of the focal point. The image can also be real when the object is placed beyond focal point.
The image produced by a diverging lens is always virtual and upright.
Thus, we can conclude that a diverging lens always produces a virtual erect image.
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