Question

A paper, with two marks having separation \[d\], is held normal to the line of sight of an observer at a distance of \[50\,{\text{m}}\] .The diameter of the eye-lens of the observer is \[2\,{\text{mm}}\] Which of the following is the least value of \[d\], so that the marks can be seen as separate? (The mean wavelength of the visible light may be taken as \[5000\,\mathop {\text{A}}\limits^{\text{o}} \]).
A. \[1.25\,{\text{m}}\]
B. \[12.5\,{\text{cm}}\]
C. \[1.25\,{\text{cm}}\]
D. \[2.5\,{\text{mm}}\]
E. \[12.5\,{\text{mm}}\]

Answer 1

Hint:Hint: Use the expression for the angular limit of the resolution of the eye. Also use the equation for the angular limit of the resolution of the eye relating the minimum just resolution separation between the two marks and their distance from the eye.

Formulae used:
The equation for the angular limit of the resolution of the eye is
\[\theta = \dfrac{\lambda }{D}\] …… (1)
Here, \[\theta \] is the angular limit of resolution of the eye, \[\lambda \] is the wavelength of the light and \[D\] is the diameter of the eye lens.
Another equation for the angular limit of the resolution of the eye is
\[\theta = \dfrac{d}{{d'}}\] …… (2)
Here, \[d\] is the minimum just resolution separation between the two marks and \[d'\] is the distance of the marks from the eye.

Complete step by step answer:
The diameter of the eye-lens is \[2\,{\text{mm}}\] and the distance of the observer from the two marks is \[50\,{\text{m}}\].The equations (1) and (2) both represent the expression for the angular limits of resolution of the eye.
Equate equations (1) and (2).
\[\dfrac{\lambda }{D} = \dfrac{d}{{d'}}\]
Rearrange the above equation for \[d\].
\[d = \dfrac{\lambda }{D}d'\]
Substitute \[5000\,\mathop {\text{A}}\limits^{\text{o}} \] for \[\lambda \], \[2\,{\text{mm}}\] for \[D\] and \[50\,{\text{m}}\] for \[d'\] in the above equation.
\[d = \dfrac{{5000\,\mathop {\text{A}}\limits^{\text{o}} }}{{2\,{\text{mm}}}}\left( {50\,{\text{m}}} \right)\]
\[ \Rightarrow d = \dfrac{{5000 \times {{10}^{ - 10}}\,{\text{m}}}}{{2 \times {{10}^{ - 3}}\,{\text{m}}}}\left( {50\,{\text{m}}} \right)\]
\[ \Rightarrow d = 12.5\,{\text{mm}}\]
Hence, the least value of \[d\]is \[12.5\,{\text{mm}}\].


Note: Convert units of all the physical quantities in the same system of units. The SI system of units is widely accepted. Hence, using proper conversion factors, convert the units of all the physical quantities in the SI system of units.