Question

A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye-lens when the eye is (a) fully relaxed, (b) most strained?

Answer 1

Given :

(a) When the eye lens is fully relaxed, we have:

u = ∞

Distance of the retina from the eye lens, v = 2 cm = 0.02 m

The lens formula is given by

1/v-1/u=1/f

Putting the  values, we get:

1/f=1/0.02-1/∞=1/0.02

∴ Power of the lens = 1/f= 50 diopters

So, in a fully relaxed condition, the power of the eye lens is 50 D.

(b) When the eye lens is most strained:

u = – 25 cm = – 0.25 m

The lens formula is given by

1/v-1/u=1/f

Putting the values, we get:

1/f=1/0.02-1/-0.25  

=50+4=54

∴ Power of the lens = 1/f= 54 diopters

So, in the most strained condition, the power of the eye lens is 54 D.

Answer 2

(ᴀ) ᴡʜᴇɴ ᴛʜᴇ ᴇʏᴇ ʟᴇɴs ɪs ғᴜʟʟʏ ʀᴇʟᴀxᴇᴅ, ᴡᴇ ʜᴀᴠᴇ:

ᴜ = ∞

ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ʀᴇᴛɪɴᴀ ғʀᴏᴍ ᴛʜᴇ ᴇʏᴇ ʟᴇɴs, ᴠ = 2 ᴄᴍ = 0.02 ᴍ

ᴛʜᴇ ʟᴇɴs ғᴏʀᴍᴜʟᴀ ɪs ɢɪᴠᴇɴ ʙʏ


1/ᴠ-1/ᴜ=1/ғ

ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ  ᴠᴀʟᴜᴇs, ᴡᴇ ɢᴇᴛ:

1/ғ=1/0.02-1/∞=1/0.02

∴ ᴘᴏᴡᴇʀ ᴏғ ᴛʜᴇ ʟᴇɴs =
1/ғ= 50 ᴅɪᴏᴘᴛᴇʀs

sᴏ, ɪɴ ᴀ ғᴜʟʟʏ ʀᴇʟᴀxᴇᴅ ᴄᴏɴᴅɪᴛɪᴏɴ, ᴛʜᴇ ᴘᴏᴡᴇʀ ᴏғ ᴛʜᴇ ᴇʏᴇ ʟᴇɴs ɪs 50 ᴅ.


(ʙ) ᴡʜᴇɴ ᴛʜᴇ ᴇʏᴇ ʟᴇɴs ɪs ᴍᴏsᴛ sᴛʀᴀɪɴᴇᴅ:

ᴜ = – 25 ᴄᴍ = – 0.25 ᴍ

ᴛʜᴇ ʟᴇɴs ғᴏʀᴍᴜʟᴀ ɪs ɢɪᴠᴇɴ ʙʏ


1/ᴠ-1/ᴜ=1/ғ

ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇs, ᴡᴇ ɢᴇᴛ:


1/ғ=1/0.02-1/-0.25  

=50+4=54

∴ ᴘᴏᴡᴇʀ ᴏғ ᴛʜᴇ ʟᴇɴs =
1/ғ= 54 ᴅɪᴏᴘᴛᴇʀs

sᴏ, ɪɴ ᴛʜᴇ ᴍᴏsᴛ sᴛʀᴀɪɴᴇᴅ ᴄᴏɴᴅɪᴛɪᴏɴ, ᴛʜᴇ ᴘᴏᴡᴇʀ ᴏғ ᴛʜᴇ ᴇʏᴇ ʟᴇɴs ɪs 54 ᴅ.