Wave Optics - Result Question 33
36. A paper, with two marks having separation $d$, is held normal to the line of sight of an observer at a distance of $50 m$. The diameter of the eye-lens of the observer is $2 mm$. Which of the following is the least value of $d$, so that the marks can be seen as separate? The mean wavelength of visible light may be taken as $5000 \AA$.
[2002]
(a) $1.25 m$
(b) $12.5 cm$
(c) $1.25 cm$
(d) $2.5 mm$
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Answer:
Correct Answer: 36. (b)
Solution:
- (b) Angular limit of resolution of eye, $\theta=\frac{\lambda}{d}$, where, $d$ is diameter of eye lens.
Also, if $Y$ is the minimum separation between two objects at distance $D$ from eye then, $\theta=\frac{Y}{D}$
$\Rightarrow \frac{Y}{D}=\frac{\lambda}{d} \Rightarrow Y=\frac{\lambda D}{d}$
Here, wavelength $\lambda=5000 \AA=5 \times 10^{-7} m$ $D=50 m$
Diameter of eye lens $=2 mm=2 \times 10^{-3} m$
From eq. (1), minimum separation is
$Y=\frac{5 \times 10^{-7} \times 50}{2 \times 10^{-3}}=12.5 \times 10^{-3} m=12.5 cm$
