Question

The diameter of the pupil of the human eye is about \[2mm\]. The human eye is most sensitive to the wavelength of \[555nm\]. The limit of resolution of the human eye is:
(A) \[1.2\text{ min}\]
(B) \[\text{2}\text{.4 min}\]
(C) \[\text{0}\text{.6 min}\]
(D) \[\text{0}\text{.3 min}\]

Answer 1

Hint: The limit of resolution (or resolving power) is a measure of the ability of the objective lens to separate in the image adjacent details that are present in the object. It is the distance between two points that are just resolved or just distinctly visible in the image.

Formula Used:
\[\delta \theta \text{=}\dfrac{1.22\lambda }{D}\]

Complete step by step answer:
We have been told the diameter of the pupil of the human eye and the wavelength of the light entering the eye. These are the only two things we need to find out the limit of resolution of the human eye.
Wavelength of light \[(\lambda )=555nm=555\times {{10}^{-9}}m\] since \[1nm={{10}^{-9}}m\]
Diameter of the lens \[(D)=2mm=2\times {{10}^{-3}}m\] since \[1mm={{10}^{-3}}m\]
Using the formula mentioned above in the hint section, we have \[\delta \theta \text{=}\dfrac{1.22\lambda }{D}\] where \[\delta \theta \] is the limit of resolution, \[\lambda \] is the wavelength of light and \[D\] is the diameter of the lens.
Substituting the values in the formula, we get
\[\begin{align}
  & \delta \theta \text{=}\dfrac{1.22\times 555\times {{10}^{-9}}}{2\times {{10}^{-3}}} \\
 & \Rightarrow \delta \theta \text{=}338.55\times {{10}^{-6}}\text{ rad} \\
\end{align}\]
Since the limit of resolution is given in radians when all the quantities are in metres (that is, in SI units), we need to convert the degree of resolution in minutes (because all the options are given in minutes).
We know that, \[\text{1 rad = 3437}\text{.75 minutes of arc}\]
Hence \[338.55\times {{10}^{-6}}\text{ rad = 3437}\text{.75}\times \text{338}\text{.55}\times \text{1}{{\text{0}}^{\text{-6}}}\text{ minutes of arc}\]
Thus degree of resolution \[\delta \theta \text{=}1.16\Rightarrow \delta \theta \simeq 1.2\text{min}\]
Hence (A) is the correct option.

Note: The actual formula for limit of resolution states that \[\text{sin}\delta \theta =\dfrac{1.22\lambda }{D}\], we neglected the trigonometric function \[\text{sin}\] here because \[\delta \theta \] is very small in magnitude, and for very small magnitude angles, sine of the angle is roughly equal to the angle itself (by approximation). This approximation stands invalid for angles of larger magnitude. Resolution is lost if the aperture of the lens is too small for the diffracted rays to be collected.