The value of numerical aperature of the objective lens of a microscope is $1.25$ . If light of wavelength $5000 ˚ A$ is used, the minimum separation between two points, to be seen as distinct, will be :

0.24 μ m

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0.48 μ m

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0.12 μ m

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0.38 μ m

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Solution

The correct option is A $0.24 μ m$
Numerical aperature of the microscope is given as
$N A = . \frac{0.61 λ}{d}$
Where $d =$ minimum sparation between two points to be seen as distinct
$d = \frac{0.61 λ}{N A} = \frac{(0.61) \times (5000 x 10 m^{- 10})}{1.25}$
$= 2.4 \times 10^{- 7} m$
$= 0.24 μ m$

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The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000˚A is used, the minimum separation between two points, to be seen as distinct, will be :

The correct option is A 0.24μmNumerical aperature of the microscope is given asNA=.0.61λdWhere d= minimum sparation between two points to be seen as distinct d=0.61λNA=(0.61)×(5000x10m−10)1.25=2.4×10−7m=0.24μm

The value of numerical aperature of the objective lens of a microscope is $1.25$ . If light of wavelength $5000 ˚ A$ is used, the minimum separation between two points, to be seen as distinct, will be :