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) $0.48 \mu m$
Show Answer
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, $\quad d=\frac{0.61 \lambda}{NA}=\frac{0.61 \times 5000 \times 10^{-10}}{1.25}$
$$ \begin{aligned} or \quad d & =\frac{3.05}{1.25} \times 10^{-7} m \\ =2.4 \times 10^{-7} m \\ or \quad d & =0.24 \mu m \end{aligned} $$
