
Diameter of human eye lens is\[2mm\]. What will be the minimum distance between two points to resolve them, which are situated at a distance of $50$ meter from eye. The wavelength of light is $5000$:-
$
A.{\text{ 2}}{\text{.32m}} \\
B.{\text{ 4}}{\text{.28m}} \\
C.{\text{ 1}}{\text{.25m}} \\
D.{\text{ 12}}{\text{.48m}} \\
$
Answer
569.7k+ views
Hint: Note down all the given known and unknown terms. Find the correlation between them and find the unknown distance from it. Convert the given terms in the same format of units. That is first convert millimetre into metre and degree Angstrom into metres and then substitute in the formula for further calculations.
Complete step by step answer:
Diameter of the human eye, $d = 2mm = 2 \times {10^{ - 3}}m$
Distance, $D = 50m$
Wavelength, $\lambda = 5000A^\circ = 5000 \times {10^{ - 10}}m$
Because $1A^\circ = {10^{ - 10}}m$
Now, let us consider “X” as the minimum distance between the two points at which lens is resolved.
Therefore, the limit of the resolution of eyeless can be given by – $1.25cm$ will be the minimum distance between two points.
$
d\theta = \dfrac{\lambda }{d} = \dfrac{x}{D} \\
\Rightarrow \dfrac{\lambda }{d} = \dfrac{x}{D} \\
$
Do-cross multiplication and make the unknown “x” as the subject –
$x = \dfrac{{\lambda D}}{d}$
Place the given values in the above equations and simplify
$x = \dfrac{{5000 \times {{10}^{ - 10}} \times 50}}{{2 \times {{10}^{ - 3}}}}$
The power of the exponents becomes positive from the negative when they are moved from denominator to the numerator and vice-versa.
$
x = \dfrac{{5000 \times {{10}^3} \times 50}}{{2 \times {{10}^{10}}}} \\
\Rightarrow x = \dfrac{{25}}{{200}} \\
\Rightarrow x = \dfrac{{1.25}}{{100}}m \\
\therefore x = 1.25cm \\
$
Therefore, the required answer - hence, from the given multiple choices- the option C is the correct answer.
Note:Always remember the units, derived units and SI units of the physical quantities to get the relation between the two or more physical quantities. Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
-MKS System (Metre Kilogram Second)
-CGS System (Centimetre Gram Second)
-System International (SI)
Also, remember the conversional relations among the system of units to make all the given units in the same format.
Complete step by step answer:
Diameter of the human eye, $d = 2mm = 2 \times {10^{ - 3}}m$
Distance, $D = 50m$
Wavelength, $\lambda = 5000A^\circ = 5000 \times {10^{ - 10}}m$
Because $1A^\circ = {10^{ - 10}}m$
Now, let us consider “X” as the minimum distance between the two points at which lens is resolved.
Therefore, the limit of the resolution of eyeless can be given by – $1.25cm$ will be the minimum distance between two points.
$
d\theta = \dfrac{\lambda }{d} = \dfrac{x}{D} \\
\Rightarrow \dfrac{\lambda }{d} = \dfrac{x}{D} \\
$
Do-cross multiplication and make the unknown “x” as the subject –
$x = \dfrac{{\lambda D}}{d}$
Place the given values in the above equations and simplify
$x = \dfrac{{5000 \times {{10}^{ - 10}} \times 50}}{{2 \times {{10}^{ - 3}}}}$
The power of the exponents becomes positive from the negative when they are moved from denominator to the numerator and vice-versa.
$
x = \dfrac{{5000 \times {{10}^3} \times 50}}{{2 \times {{10}^{10}}}} \\
\Rightarrow x = \dfrac{{25}}{{200}} \\
\Rightarrow x = \dfrac{{1.25}}{{100}}m \\
\therefore x = 1.25cm \\
$
Therefore, the required answer - hence, from the given multiple choices- the option C is the correct answer.
Note:Always remember the units, derived units and SI units of the physical quantities to get the relation between the two or more physical quantities. Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
-MKS System (Metre Kilogram Second)
-CGS System (Centimetre Gram Second)
-System International (SI)
Also, remember the conversional relations among the system of units to make all the given units in the same format.
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