Question

A person can not see an object lying beyond \[80\,cm\], whereas a normal person can easily see the object kept at a distance of \[160\,cm\], the focal length and nature of the lens used to rectify this defect will be?
A. \[160\,cm\], concave
B. \[160\,cm\], convex
C. \[60\,cm\], concave
D. \[60\,cm\], convex

Answer 1

Hint: To solve this question we have to know about concave and convex lenses. We know that a concave lens is a lens that has in any event one surface that bends inwards. It is a veering focal point, implying that it fans out light beams that have been refracted through it.

Complete step by step answer:
We know that a convex lens or meeting focal point shines the light beams to a particular point while a curved focal point or wandering focal point separates the light beams.
to diverge the rays, concave lenses will be used. We know that,
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
Here $v$ is the length between image and the centre of the lens. We know that a focal point is a straightforward material (either bended or level surface) in light of the standards of refraction. $u$ is the length between the object and the center of the lens and $f$ is the focal length. Putting the values we will get,
\[\dfrac{1}{{160}} - \dfrac{1}{{80}} = \dfrac{1}{f}\]
\[ \therefore f = - 160\,cm\]

Since a concave lens has a negative focal length thus, option A is correct.

Note: We know that, since a concave lens will not produce a real image, a convex lens is used to measure its focal length. We have to keep these in our mind to solve questions. At the point when an item is set at a limited separation from the focal point, a virtual picture is framed among the shaft and focal point of the raised focal point. The size of the picture is more modest when contrasted with that of the item.