For a light of wavelength 250 nm, calculate the resolving power of a telescope whose objective lens has an aperture of 0.5 m.

Question

Question

For a light of wavelength 250 nm, calculate the resolving power of a telescope whose objective lens has an aperture of 0.5 m.

\(3.28\times10^{6}\)
\(1.64\times10^{6}\)
\(6.4\times10^{6}\)
\(1.64\times10^{-6}\)

Answer 1

Concept:

Resolving power is defined as the inverse of the distance or angular separation between two objects which can be just resolved when viewed through the optical instruments.

According to Rayleigh criterion

In telescope

angular separation b/w two object

\({\rm{\Delta }}\theta = 1.22\frac{\lambda }{d}\)

Resolving power \( = \frac{1}{{{\rm{\Delta }}\theta }} = \frac{d}{{1.22\lambda }}\)

d = diameter

λ = wavelength

In microscope

Distance between two object

\({\rm{\Delta }}d = \frac{\lambda }{{2n\sin \theta }}\)

Resolving power \( = \frac{1}{{{\rm{\Delta }}d}} = \frac{{2n\sin \theta }}{\lambda }\)

n = refractive index of medium separating object and aperture

Calculations:

Given,

Wavelength = l = 250 nm = 2.5 × 10^-7m

Aperture = D = 0.5 m

We know,

The resolving power of the telescope =

\(\frac{D}{1.22\times \lambda}=\frac{0.5}{1.22\times 2.5 \times 10^{-7}}=1.64\times10^{6}\)

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For a light of wavelength 250 nm, calculate the resolving power of a telescope whose objective lens has an aperture of 0.5 m.

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