Question

A far sighted man cannot focus distinctly on objects closer than 1 metre. The power of the lens that will permit him to read from a distance of 40 cm is?

Answer 1

HEY DEAR ,
HERE IS YOUR ANSWER ,

==>
focal length is given as

1/f = 1/v - 1/u

here

u = -40 cm

v = -100 cm

so,

1/f = (1/-100) - (1/-40)

1/f = (-2+5)/200

1/f = 3/200

or focal length

f = 200/3 cm = 66.67 cm

and power of lens

[here f = 66.67 cm = 0.66m]

P = 1/f = 1/0.66= 1.51 D

☺

Answer 2

Answer:

Concept:

There are numerous causes of eye defects. As we become older, our eyesight also gets worse, and when our focus length changes, so do our vision. We are aware that the most typical eye defect is a cataract. When left untreated, cataracts can result in a partial loss of eyesight or even total blindness. A cataract is when the crystalline lens ages and becomes milky and cloudy. A person's vision can be recovered after cataract surgery.

Step-by-step explanation:

When the eye loses the capacity to change its focal length, issues arise such as blurred vision, an inability to perceive close-up or distant objects, and trouble seeing the image clearly. The user is unable to clearly and comfortably view the items when the refractive index problem occurs. Without prompt attention, eyes may completely lose their ability to accommodate. Learn about numerous visual problems and how to correct them in this post.

                    $Near \ point, \mathrm{v}=-100 \mathrm{~cm} \\\\Reading Distance, \mathrm{u}=-40 \mathrm{~cm} \\\\From \ lens \ formula,\frac{1}{f}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\\\\ =\frac{1}{-100}-\frac{1}{-40}=\frac{3}{200}\\\mathrm{f}=0.015 \mathrm{~cm}\\Focal length of lens f= 0.015 \mathrm{~cm}$

A far-sighted man cannot focus distinctly on objects closer than 1 meter. The power of the lens that will permit him to read from a distance of 40 cm is 0.015cm.

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