Wave Optics Ques 30
30. 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: 30.(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, $\theta=\frac{Y}{D}$
$\Rightarrow \frac{Y}{D}=\frac{\lambda}{d} \Rightarrow Y=\frac{\lambda D}{d}$ $\quad$ …….(1)
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$
BETA
