
A person needs a lens of power -5.5 diopters for correcting his distant vision. For correcting his near vision he needs a lens of power 1.5+ diopters. What is the focal length of the lens required for correcting (i) distant vision (ii) near vision?
Answer
514.8k+ views
Hint- In order to solve these types of questions, we will use the concept of power of the lens and we will proceed further first by calculating the focal length using the power and then proceed further for finding the distance of distant vision and near vision.
Complete step-by-step answer:
The formula for power of lens is given by \[P = \dfrac{1}{f}\]
Where P is the power and f is the focal length.
(i) The power of the lens required for correcting the vision = – 5.5 D
Focal length of the lens,
\[
f = \dfrac{1}{P} \\
f = \dfrac{1}{{ - 5.5}}{\text{ }} \\
= - 0.181{\text{ }}m \\
\]
Therefore the required focal length for correcting the vision is $ - 0.181m$
(ii) The power of the lens required for correcting near vision = +1.5 D.
Focal length of the lens is \[f = \dfrac{1}{P}\]
Therefore, the focal length of the lens for correcting near vision is 0.667 m
Additional Information- Lenses can be classified into two major forms of large measure: convex and concave. The lenses in the middle are smoother than the corners, while the lenses along the sides are concave. The mirror concentrates on a spot on the other side of the frame, a light ray traveling through a convex prism. This is considered the focus of publicity. In the case of a concave lens that diverges from the condensation of light waves, the focal point is on the source of incoming light that appears to come from the direction of light through the lens.
Note- In order to solve these types of questions, apart from the formulas you must remember the sign conventions. Some of the sign conventions are- for a convex lens the focal length is positive and for a concave lens the focal length is negative. Similarly others remember the other conventions. Most optical devices make use of not just one lens, but of a combination of convex and concave lenses.
Complete step-by-step answer:
The formula for power of lens is given by \[P = \dfrac{1}{f}\]
Where P is the power and f is the focal length.
(i) The power of the lens required for correcting the vision = – 5.5 D
Focal length of the lens,
\[
f = \dfrac{1}{P} \\
f = \dfrac{1}{{ - 5.5}}{\text{ }} \\
= - 0.181{\text{ }}m \\
\]
Therefore the required focal length for correcting the vision is $ - 0.181m$
(ii) The power of the lens required for correcting near vision = +1.5 D.
Focal length of the lens is \[f = \dfrac{1}{P}\]
Therefore, the focal length of the lens for correcting near vision is 0.667 m
Additional Information- Lenses can be classified into two major forms of large measure: convex and concave. The lenses in the middle are smoother than the corners, while the lenses along the sides are concave. The mirror concentrates on a spot on the other side of the frame, a light ray traveling through a convex prism. This is considered the focus of publicity. In the case of a concave lens that diverges from the condensation of light waves, the focal point is on the source of incoming light that appears to come from the direction of light through the lens.
Note- In order to solve these types of questions, apart from the formulas you must remember the sign conventions. Some of the sign conventions are- for a convex lens the focal length is positive and for a concave lens the focal length is negative. Similarly others remember the other conventions. Most optical devices make use of not just one lens, but of a combination of convex and concave lenses.
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