Optics 5 Question 1
1. 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: 1. (a)
Solution:
- 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
$$ \begin{aligned} d & =\frac{3.05}{1.25} \times 10^{-7} m \\ & =2.4 \times 10^{-7} m \\ d & =0.24 \mu m \end{aligned} $$
or
2 Limit of resolution for a telescope from Rayleigh’s criteria is
$$ \theta _R=\frac{1.22 \lambda}{D} $$
Here, $\quad D=250 cm=250 \times 10^{-2} m$
and $\quad \lambda=600 nm=600 \times 10^{-9} m$
So, limit of resolution is
$$ \begin{aligned} \theta _R & =\frac{1.22 \times 600 \times 10^{-9}}{250 \times 10^{-2}} \\ & =2.93 \times 10^{-7} rad \\ & \approx 3 \times 10^{-7} rad \end{aligned} $$
3 For a telescope, limit of resolution is
$$ \text { given by } \Delta \theta=\frac{1.22 \lambda}{D} $$
Here, $\lambda=500 nm=500 \times 10^{-9} m$,
$$ D=200 cm=200 \times 10^{-2} m $$
So, limit of resolution is $\Delta \theta=\frac{1.22 \times 500 \times 10^{-9}}{200 \times 10^{-2}}$
$$ =305 \times 10^{-9} rad $$
