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The head lights of a jeep are $1.2 \mathrm{~m}$ apart. If the pupil of the eye of an observer has a diameter of $2 \mathrm{~mm}$ and light of wavelength $5896 Å$ is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated?
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Verified Answer
The correct answer is:
$3.34 \mathrm{~m}$
Distance of jeep, $x=\frac{D \times d}{1.22 \times \lambda}$
where $D=$ diameter of lens,
$$
\begin{aligned}
d &=\text { seperation between sources } \\
\Rightarrow \quad x &=\frac{\left(2 \times 10^{-3}\right) \times 1.2}{1.22 \times 5896 \times 10^{-10}}=3336 \mathrm{~m} \\
\Rightarrow \quad x &=3.34 \mathrm{~km}
\end{aligned}
$$
where $D=$ diameter of lens,
$$
\begin{aligned}
d &=\text { seperation between sources } \\
\Rightarrow \quad x &=\frac{\left(2 \times 10^{-3}\right) \times 1.2}{1.22 \times 5896 \times 10^{-10}}=3336 \mathrm{~m} \\
\Rightarrow \quad x &=3.34 \mathrm{~km}
\end{aligned}
$$
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