# Or a Normal Eye, the Far Point is at Infinity and the Near Point of Distinct Vision is About 25cm in Front of the Eye. from this Rough Data Estimate the Range of Accommodation (I.E., the Range of Converging Power of the Eye-lens) of a Normal Eye. - Physics

For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.

#### Solution

Least distance of distinct vision, d = 25 cm

Far point of a normal eye, d' =oo

Converging power of the cornea, P_c =40D

Least converging power of the eye-lens, P_c = 20D

To see the objects at infinity, the eye uses its least converging power.

Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60 D

Power of the eye-lens is given as: P = 1/"Focal length of the eye lens(f)"

f = 1/P

= 1/60D

= 100/60 = 5/3 cm

To focus an object at the near point, object distance (u) = −d = −25 cm

Focal length of the eye-lens = Distance between the cornea and the retina

= Image distance

Hence, image distance, v = 5/3 cm

According to the lens formula, we can write:

1/f' = 1/v - 1/u

Where,

f' = Focal length

1/f' = 3/2 + 1/25 = (15+1)/25 = 16/25 cm

Power, P' = 1/"f'" xx 100

= 16/25 xx 100 = 64D

∴Power of the eye-lens = 64 − 40 = 24 D

Hence, the range of accommodation of the eye-lens is from 20 D to 24 D.

Concept: Optical Instruments - The Eye
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#### APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 24 | Page 348
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