The semivertical angle of cone of the rays incident on the objective of microscope is $20^{0}$ . If the wavelength of incident of light ray is 6000 A, calculate the smallest distance between two points which can be just resolved.

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Solution

For the microscope
semivertical angle $= 20^{o} = θ$
and $λ = 6 \times 10^{- 7} m$
thus angular resolving power of microscope
$θ_{R P} = \frac{λ}{2 μ sin θ}$
$μ =$ Refractive index of gain $= 3 / 2$
$= θ_{R P} = \frac{2}{2 \times 3 \times s i n (20^{o})} = 0.975$ rad

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The semivertical angle of cone of the rays incident on the objective of microscope is 200. If the wavelength of incident of light ray is 6000 A, calculate the smallest distance between two points which can be just resolved.

For the microscopesemivertical angle =20o=θand λ=6×10−7mthus angular resolving power of microscope θRP=λ2μsinθμ= Refractive index of gain =3/2=θRP=22×3×sin(20o)=0.975 rad

The semivertical angle of cone of the rays incident on the objective of microscope is $20^{0}$ . If the wavelength of incident of light ray is 6000 A, calculate the smallest distance between two points which can be just resolved.

For the microscope
semivertical angle $= 20^{o} = θ$
and $λ = 6 \times 10^{- 7} m$
thus angular resolving power of microscope
$θ_{R P} = \frac{λ}{2 μ sin θ}$
$μ =$ Refractive index of gain $= 3 / 2$
$= θ_{R P} = \frac{2}{2 \times 3 \times s i n (20^{o})} = 0.975$ rad