JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 15)
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 :
0.12 $$\mu $$m
0.38 $$\mu $$m
0.24 $$\mu $$m
0.48 $$\mu $$m
Explanation
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 $$ \times {10^{ - 7}}\,m = 0.24\,\mu m$$
$$NA = {{0.61\lambda } \over d}$$
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 $$ \times {10^{ - 7}}\,m = 0.24\,\mu m$$
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