
Concave lens always forms
A. Virtual image
B. Erect image
C. Diminished image
D. All of the above
Answer
507k+ views
Hint:Concave lenses have one or more surfaces that curve inwards. They are diverging lenses, meaning that they spread light rays refracted through them. They are used for the treatment of myopia as they make faraway objects look smaller than they really are.
Complete answer:
As we know, objects are kept behind the lens so the value of $u$ will be negative and the focus of the concave lens is negative. So, by using Lens formula
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
We can find value of $v$ =image distance
So, the value of $v$ is negative as the image is formed behind the lens.We will draw tables for the nature of the image formed by a lens.
Here if the object is very far away, the images formed by lenses will be all the more diminished.So, by we get the image is always formed between lens and object instead of whatever is position of the object. By using magnification formula:
\[m = \dfrac{{{h_i}}}{{{h_0}}} = \dfrac{v}{u}\]
We get an image size that is less than the object and it is diminished in size. In case of the concave lens, we always get erect images, diminished images and virtual images
Hence. option D is correct.
Note:Lenses are classified according to the curvature of both the optic surfaces. Only if both surfaces are convex, will a lens be referred to as biconvex. If one side of the lens is flat, then the lens can be either plano concave or plano convex depending on the curvature of the other side. A lens with two concave surfaces is considered to be double concave, or meniscus. Corrective lenses of this kind are commonly used.
Complete answer:
As we know, objects are kept behind the lens so the value of $u$ will be negative and the focus of the concave lens is negative. So, by using Lens formula
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
We can find value of $v$ =image distance
So, the value of $v$ is negative as the image is formed behind the lens.We will draw tables for the nature of the image formed by a lens.
| Object location | Image location | Image nature | Image size |
| Infinity | At ${F_2}$ | Virtual and Erect | Highly Diminished |
| Beyond infinity and 0 | Between ${F_1}$ and Optical centre | Virtual and Erect | Diminished |
Here if the object is very far away, the images formed by lenses will be all the more diminished.So, by we get the image is always formed between lens and object instead of whatever is position of the object. By using magnification formula:
\[m = \dfrac{{{h_i}}}{{{h_0}}} = \dfrac{v}{u}\]
We get an image size that is less than the object and it is diminished in size. In case of the concave lens, we always get erect images, diminished images and virtual images
Hence. option D is correct.
Note:Lenses are classified according to the curvature of both the optic surfaces. Only if both surfaces are convex, will a lens be referred to as biconvex. If one side of the lens is flat, then the lens can be either plano concave or plano convex depending on the curvature of the other side. A lens with two concave surfaces is considered to be double concave, or meniscus. Corrective lenses of this kind are commonly used.
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