Question

The diameter of the eyeball is about 2.5cm. The power of the eye lens varies from
A. 9 D to 8 D
B. 40 D to 32 D
C. 44 D to 40 D
D. None of these

Answer 1

Hint: Basic knowledge of the eye is necessary to solve this problem. That is how far are we able to see things in real life easily and by straining our eyes. The same condition for distant objects too. We will also be using the lens makers formula, ${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$ , since the eye is a lens. Further, power of a lens is given by inverse of the focal length of the lens, that is $P={\displaystyle \frac{1}{f}}.$

Complete step-by-step answer:
So, here we will have to consider a real life scenario to solve this problem. For a healthy eye, there are a few conditions given to have a normal comfortable vision. These are, a normal healthy eye can see very distant objects comfortably without having any strain on the eye and similarly a healthy eye can also view objects as close upto 25cm close to the eye comfortably, without any strain on the eye.
 $\to$ Let’s consider the case, when we see distant objects comfortably.
Here the object distance will be considered infinity, that is $u=-\mathrm{\infty }$ and the image distance will be at the diameter of the eye lens, that is $v=2.5cm\Rightarrow v=2.5\times {10}^{-2}m.$
Then using the basic formula of optics binding u,v and f, that is:
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
Substituting the values of this case into the above formula, we get:
${\displaystyle \frac{1}{2.5\times {10}^{-2}}}-{\displaystyle \frac{1}{\mathrm{\infty }}}={\displaystyle \frac{1}{f}}$
We know that power (P) is equal to the inverse of focal length, that is:
$P={\displaystyle \frac{1}{f}}.$
Hence, for this case, $\because {\displaystyle \frac{1}{\mathrm{\infty }}}=0,$
$\therefore {\displaystyle \frac{1}{2.5\times {10}^{-2}}}={\displaystyle \frac{1}{f}}=P\Rightarrow P=40D.$
 $\to$ Now, we will consider the case when we see objects at 25cm distance straining our eyes.
Here, the object is at 25cm distance, that is $u=-25cm\Rightarrow u=-25\times {10}^{-2}m$ and the image distance will be at the diameter of the eye lens, that is $v=2.5cm\Rightarrow v=2.5\times {10}^{-2}m.$
Using the same formula as above, $P={\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$
Substituting in the values of current case into the equation, we get: $P={\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{2.5\times {10}^{-2}}}-{\displaystyle \frac{1}{-25\times {10}^{-2}}}\Rightarrow P=40D+4D\Rightarrow P=44D.$
Hence, the power of the eye lens varies from 40D to 44D.

Note: A general question that you may ask is, why is the object distance taken as negative. I’ve considered $u=-\mathrm{\infty }$ in the first case and $u=-25cm$ in the second case. This is because of the sign convention followed in general. In optics, all the rays coming from left of the optical center are taken as negative and all the rays to the right of the optical center are considered positive.
To remember it easily, you can consider the optical center to be the zero of a number line, where the object is to the left or the negative of the number line and the image is formed to the right of the optical center or the positive part of the number line.