Optics Ques 121
- The value of numerical aperture of the objective lens of a microscope is $1.25$ . If light of wavelength $5000$ $ \AA$ is used, the minimum separation between two points, to be seen as distinct, will be
(2019 Main, 12 April I)
(a) $0.24 $ $\mu m$
(b) $0.38 $ $\mu m$
(c) $0.12 $ $\mu m$
(d) $0.48$ $ \mu m$
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Answer:
Correct Answer: 121.(a)
Solution:
Formula:
- Key Idea: Numerical Aperture (NA) of the microscope is given by $N A=\frac{0.61 \lambda}{d}$
Here, $d=$ minimum separation between two points to be seen as distinct and $\lambda$ =wavelength of light.
Given, $\lambda=5000 $ $\AA=5000 \times 10^{-10} $ $m$ and
$NA=1.25$
Now, $\quad d=\frac{0.61 \lambda}{NA}=\frac{0.61 \times 5000 \times 10^{-10}}{1.25}$
or
$ \begin{aligned} d & =\frac{3.05}{1.25} \times 10^{-7} m \\ & =2.4 \times 10^{-7} m \\ d & =0.24 \mu m \end{aligned} $
BETA
