QJEEM19Ph5b33: JEE Main 2019 Physics - Class 12 JEEM19Ph5b33 Solution: The value of numerical aperature of the objective lens of a…

Question

Verified Solution • 2025–26 Session moderate

The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000

\mathop A\limits^o

is used, the minimum separation between two points, to be seen as distinct, will be :

Accepted Answer

Numerical aperature of the microscope is given as

NA = {{0.61\lambda } \over d}

Where d = minimum sparaton between two points to be seen as distinct

d = {{0.61\lambda } \over {NA}} = {{\left( {0.61} \right) \times \left( {5000 \times 10\,{m^{ - 10}}} \right)} \over {1.25}}

= 2.4

\times {10^{ - 7}}\,m = 0.24\,\mu m

Step-by-Step Verified Solution