Question

The near and far points of a person are at 40 cm and 250 cm respectively. With aided eye, the maximum distance upto which he can clearly see is approximately _____ cm (given: Focal length of the lens used +200/3 cm)​

Answer 1

Thus, u=−25cm,v=−40cm

1/f=1/v−1/u

=1/40cm+1/25cm

or f=200/3cm=+23/m

P=1/f+1.5

1/v−1/u=1/f

or−1/250cm−1/d=3/200cm

or d=−53cm

Answer 2

Given :

Near point of person = 40 cm

The far point of person = 250 cm

The focal length of lens = $\frac{+200}{3}$ cm

To Find:

Maximum distance up to which he can see with aided eye.

Solution :

• The far point of the human eye is the maximum distance till which the eye can distinguish objects clearly.
• For a normal eye, the far point is at infinity. This can recede due to some defects of the eye such as myopia.
• We will use the lens formula for solving this question, which is:

                                            $\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$

where  v is the distance of the image formed

            u is the distance of the object

     and f is the focal length of the lens

• So, for finding maximum distance up to which the person can see, we have the following values:

1. The focal length of lens = $\frac{+200}{3}$ cm
2. Let maximum distance up to which he can see be d cm
3. The distance at which image is formed when the object is kept at d cm = 250 cm

• Applying the lens formula:

             ⇒  $\frac{-1}{250}$ - ($\frac{-1}{d}$) = $\frac{3}{200}$

             ⇒  $\frac{1}{d}$ = $\frac{3}{200}$ + $\frac{1}{250}$

             ⇒   $\frac{1}{d}$ = $\frac{19}{1000}$

             ⇒   d = $\frac{1000}{19}$ cm ≈ 53cm.

Therefore, with the aided eye, the maximum distance up to which he can see is approximately 53 cm.