KCET · Physics · Ray Optics
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?
- A \(33.9 \mathrm{~km}\)
- B \(33.9 \mathrm{~m}\)
- C \(3.34 \mathrm{~m}\)
- D \(3.39 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(3.34 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
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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