Answer
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Hint: We are given the wavelength of light to which the human eye is most sensitive and the diameter of the pupil. We are asked to find the angular limit of the resolution of the eye. We have an equation to find the angular limit of the resolution of the eye. By substituting the given values in the known equation we will get the required solution.
Formula used:
${{D}_{\theta }}=\dfrac{1.22\lambda }{d}$
Complete step-by-step solution
In the question, it is said that the human eye is most sensitive to light with wavelength $5500\overset{o}{\mathop{A}}\,$. We are also given the diameter of the pupil as 2 mm.
Therefore we gave,
Wavelength, $\lambda =5500\overset{o}{\mathop{A}}\,$
Since this is given in angstrom, we can convert it into meters.
Thus we get,
Wavelength, $\Rightarrow \lambda =5500\times {{10}^{-10}}m$
And the diameter of the pupil,
$d=2mm$
By converting this into meters, we get
$d=2\times {{10}^{-3}}m$
We are asked to calculate the angular limit of the resolution of the eye.
We know that the equation for an angular limit of resolution for the eye is given as,
${{D}_{\theta }}=\dfrac{1.22\lambda }{d}$, were ‘$\lambda $’ is the wavelength and ‘d’ is the diameter
By substituting the values for wavelength and diameter in the above equation, we will get
\[{{D}_{\theta }}=\dfrac{1.22\times 5500\times {{10}^{-10}}}{2\times {{10}^{-3}}}\]
By solving this we will get,
$\Rightarrow {{D}_{\theta }}=3.355\times {{10}^{-4}}rad$
Therefore the angular limit of resolution of eye is $3.355\times {{10}^{-4}}radians$.
Note: Angular limit of resolution or resolving power is simply the ability of an eye or an optical instrument to differentiate between two objects or points. The sensitivity of the human eye to light varies over lights with its wavelength 380 – 800 nm. But in normal daylight conditions, the human eye is most sensitive to light of wavelength 555 nm.
Formula used:
${{D}_{\theta }}=\dfrac{1.22\lambda }{d}$
Complete step-by-step solution
In the question, it is said that the human eye is most sensitive to light with wavelength $5500\overset{o}{\mathop{A}}\,$. We are also given the diameter of the pupil as 2 mm.
Therefore we gave,
Wavelength, $\lambda =5500\overset{o}{\mathop{A}}\,$
Since this is given in angstrom, we can convert it into meters.
Thus we get,
Wavelength, $\Rightarrow \lambda =5500\times {{10}^{-10}}m$
And the diameter of the pupil,
$d=2mm$
By converting this into meters, we get
$d=2\times {{10}^{-3}}m$
We are asked to calculate the angular limit of the resolution of the eye.
We know that the equation for an angular limit of resolution for the eye is given as,
${{D}_{\theta }}=\dfrac{1.22\lambda }{d}$, were ‘$\lambda $’ is the wavelength and ‘d’ is the diameter
By substituting the values for wavelength and diameter in the above equation, we will get
\[{{D}_{\theta }}=\dfrac{1.22\times 5500\times {{10}^{-10}}}{2\times {{10}^{-3}}}\]
By solving this we will get,
$\Rightarrow {{D}_{\theta }}=3.355\times {{10}^{-4}}rad$
Therefore the angular limit of resolution of eye is $3.355\times {{10}^{-4}}radians$.
Note: Angular limit of resolution or resolving power is simply the ability of an eye or an optical instrument to differentiate between two objects or points. The sensitivity of the human eye to light varies over lights with its wavelength 380 – 800 nm. But in normal daylight conditions, the human eye is most sensitive to light of wavelength 555 nm.
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