Question

The far point of a myopia person is 80cm in front of the eye what is the nature and power of the lens required to correct the defect.​

Answer 1

Answer

The defect called Myopia is correct by using a concave lens. So, the person requires concave lens spectacles.

Firstly we have to find focal length of the concave lens required.

According to the question,

The image distance of the is 80cm and the object distance is infinitely.

now, using lens formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

$\boxed{ \bf \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} }$

where,

• v denotes image distance
• u denotes object distance
• f denotes focal length

by substituting all the given values in the formula,

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} }$

$\leadsto{ \sf \dfrac{1}{ - 80} - \dfrac{1}{ \infty } = \dfrac{1}{f} }$

$\leadsto{ \sf \dfrac{1}{ (- 8)} - 0 = \dfrac{1}{f} }$

$\leadsto{ \sf \dfrac{1}{ - 80} = \dfrac{1}{f} }$

$\leadsto{ \sf f = - 80 \: cm}$

thus, the focal length of the required lens is -80cm

Now, we have to find the Power of the lens,

» The power of a lens is defined as the reciprocal of its focal length in metres.

f = - 80cm = 0.8m.

$\leadsto{ \sf P = \dfrac{1}{f} }$

$\leadsto{ \sf P = - \dfrac{1}{0.8} }$

$\leadsto{ \sf P = - \dfrac{10}{8} }$

$\leadsto{ \sf P = - 1.25 \: D}$

thus, the power of the lens is - 1.25 D