Answer
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Hint:In order to answer this question we should first know about lense. Lenses are essentially curved-side magnifying glasses. By refraction, a lens is a piece of clear glass that concentrates or disperses light rays when they move through it. Lenses are used in telescopes and other magnifying instruments because of their magnifying properties. They are used to collect light rays and are used in cameras.
Complete answer:
Now we'll look at the lens formula. In optics, the Lens formula defines the relationship between the image distance (v), the object distance (u), and the focal length (f) of the lens.Both convex and concave lenses are compatible with the lens formula. These lenses have a very thin thickness.
The following is the formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$, where \[v\] is the image distance, \[u\] is the object distance, and \[f\] is the lens focal length. The optical centre of the lens is used to determine the distance between the object and the image. The distance sign is given according to convention.
So, the correct option is C.
Note: The lens formula is applicable to both convex and concave lenses. It can also be used to figure out how far actual and virtual images are apart. If the equation yields a negative image distance, the resulting image is virtual and on the same side as the object.
Complete answer:
Now we'll look at the lens formula. In optics, the Lens formula defines the relationship between the image distance (v), the object distance (u), and the focal length (f) of the lens.Both convex and concave lenses are compatible with the lens formula. These lenses have a very thin thickness.
The following is the formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$, where \[v\] is the image distance, \[u\] is the object distance, and \[f\] is the lens focal length. The optical centre of the lens is used to determine the distance between the object and the image. The distance sign is given according to convention.
So, the correct option is C.
Note: The lens formula is applicable to both convex and concave lenses. It can also be used to figure out how far actual and virtual images are apart. If the equation yields a negative image distance, the resulting image is virtual and on the same side as the object.
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