At night ,on a long stretch of straight road in Kota, a truck driver sees the distant headlights of another truck. How close must he be to the other truck in order for his eyes to just resolve two headlights? Assume that the pupils of the truck driver have a diameter of 5mm, that the headlights are separated by $1.22\,m$, and that the light has wavelength $500\,nm$.
A. 10 km
B. 1 km
C. 30 km
D. 100 km
Answer
533.7k+ views
Hint: to solve this question we will first start by finding the angular resolving power. later we will find the distance . We need to find resolving power angle first because to find the distance of driver from headlights we use a formula that relates resolving power and separated distance(l)
Formula used:
\[\theta = \dfrac{{1.22\lambda }}{d}\]
Where, \[\theta \] is the angular resolving power, \[\lambda \] is the wavelength and \[d\] is the diameter of the pupil of the eye.
\[\theta = \dfrac{L}{D}\]
Where, \[\theta \]is the angular resolving power,\[L\]- distance between 2 headlights and \[D\]- the distance of driver from headlights
Complete step by step answer:
Let us look at the definition of resolving power: it is defined as the ability of an optical instrument or type of film to separate or distinguish small or closely adjacent images.
Given: \[d = 5\,mm = 5 \times {10^{ - 3}}\,m\], \[L = 1.22\,m\] and \[\lambda = 500 \times {10^{ - 9}}\,m\].
Let us start by first finding \[\theta \],
\[\theta = \dfrac{{1.22\lambda }}{d}\]
\[ \Rightarrow \theta = \dfrac{{1.22 \times 500 \times {{10}^{ - 9}}}}{{5 \times {{10}^{ - 3}}}}\]
\[ \Rightarrow \theta = \dfrac{{1.22 \times 500 \times {{10}^{ - 9 + 3}}}}{5}\]
\[ \Rightarrow \theta = \dfrac{{1.22 \times 500 \times {{10}^{ - 6}}}}{5}\]
\[ \Rightarrow \theta = 122 \times {10^{ - 6}}\]
\[ \Rightarrow \theta = 1.22 \times {10^{ - 4}}\]
Now we will find the value of D.
\[\theta = \dfrac{L}{D}\]
\[\Rightarrow D = \dfrac{L}{\theta }\]
\[\Rightarrow D = \dfrac{{1.22}}{{1.22 \times {{10}^{ - 4}}}}\]
\[\Rightarrow D = {10^4}m\]
\[\therefore D = 10\,km\].
Hence the distance of the driver from headlights is 10 km.
Hence the correct answer is option A.
Note:Remember that L is the given distance between any two objects but $D$ is always the distance between source and the object/ body. Hence we have to find the value of $D$ not $L$. $L$ is already given to be 1.22 m. Resolving power should not be given a unit of meter, it is unitless.
Formula used:
\[\theta = \dfrac{{1.22\lambda }}{d}\]
Where, \[\theta \] is the angular resolving power, \[\lambda \] is the wavelength and \[d\] is the diameter of the pupil of the eye.
\[\theta = \dfrac{L}{D}\]
Where, \[\theta \]is the angular resolving power,\[L\]- distance between 2 headlights and \[D\]- the distance of driver from headlights
Complete step by step answer:
Let us look at the definition of resolving power: it is defined as the ability of an optical instrument or type of film to separate or distinguish small or closely adjacent images.
Given: \[d = 5\,mm = 5 \times {10^{ - 3}}\,m\], \[L = 1.22\,m\] and \[\lambda = 500 \times {10^{ - 9}}\,m\].
Let us start by first finding \[\theta \],
\[\theta = \dfrac{{1.22\lambda }}{d}\]
\[ \Rightarrow \theta = \dfrac{{1.22 \times 500 \times {{10}^{ - 9}}}}{{5 \times {{10}^{ - 3}}}}\]
\[ \Rightarrow \theta = \dfrac{{1.22 \times 500 \times {{10}^{ - 9 + 3}}}}{5}\]
\[ \Rightarrow \theta = \dfrac{{1.22 \times 500 \times {{10}^{ - 6}}}}{5}\]
\[ \Rightarrow \theta = 122 \times {10^{ - 6}}\]
\[ \Rightarrow \theta = 1.22 \times {10^{ - 4}}\]
Now we will find the value of D.
\[\theta = \dfrac{L}{D}\]
\[\Rightarrow D = \dfrac{L}{\theta }\]
\[\Rightarrow D = \dfrac{{1.22}}{{1.22 \times {{10}^{ - 4}}}}\]
\[\Rightarrow D = {10^4}m\]
\[\therefore D = 10\,km\].
Hence the distance of the driver from headlights is 10 km.
Hence the correct answer is option A.
Note:Remember that L is the given distance between any two objects but $D$ is always the distance between source and the object/ body. Hence we have to find the value of $D$ not $L$. $L$ is already given to be 1.22 m. Resolving power should not be given a unit of meter, it is unitless.
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