SATHEE: Optics 5 Question 1

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) $0.24 μ m$

(b) $0.38 μ m$

(d) $0.48 μ m$

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Answer:

Correct Answer: 1. (a)

Solution:

Key Idea

Numerical Aperture (NA) of the microscope is given by $N A = \frac{0.61 λ}{d}$

Here, $d =$ minimum separation between two points to be seen as distinct and $λ$ =wavelength of light.

Given, $λ = 5000 \AA = 5000 \times 10^{- 10} m$ and $N A = 1.25$

Now, $d = \frac{0.61 λ}{N A} = \frac{0.61 \times 5000 \times 10^{- 10}}{1.25}$

$\begin{aligned} o r d & = \frac{3.05}{1.25} \times 10^{- 7} m \\ = 2.4 \times 10^{- 7} m \\ o r d & = 0.24 μ m \end{aligned}$

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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

Answer:

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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