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Question: The angular resolution of a telescope of \(10cm\) diameter at a wavelength of \(\lambda =5000\overse...

The angular resolution of a telescope of 10cm10cm diameter at a wavelength of λ=5000A0 \lambda =5000\overset{0}{\mathop{A}}\, is of the order of ………
A.106rad B.10−2rad C.10−4rad D.10−6rad \begin{aligned} & A{{.10}^{6}}rad \\\ & B{{.10}^{-2}}rad \\\ & C{{.10}^{-4}}rad \\\ & D{{.10}^{-6}}rad \\\ \end{aligned}

Explanation

Solution

The angular resolution or also called as angle of resolving power of a telescope is described as the smallest angle between the object which can be seen perfectly as separate. The wave nature of light limits the angle of resolving power. Two point sources are considered to be as resolved if the first diffraction maximum of one of the images coincides with the principal minimum of the other.
Formula used:
θ=1.22λd\theta =1.22\dfrac{\lambda }{d}
Where θ\theta be the angle of resolution, λ\lambda be the wavelength of the light used, and dd be the diameter of lens aperture.

Complete answer:
If we are considering this through a circular aperture, it can be translated to an expression which is given as,
θ=1.22λd\theta =1.22\dfrac{\lambda }{d}
Where θ\theta be the angle of resolution, λ\lambda be the wavelength of the light used, and dd be the diameter of lens aperture of the telescope.
The value of the terms given in the question can be written as,
The wavelength of the light used is,
λ=5000A0 \lambda =5000\overset{0}{\mathop{A}}\,
The angstrom should be converted into metre. That is,
⇒λ=5000A0 =5000×10−10m\Rightarrow \lambda =5000\overset{0}{\mathop{A}}\,=5000\times {{10}^{-10}}m λ=5000A0 =5000×10−10m\lambda =5000\overset{0}{\mathop{A}}\,=5000\times {{10}^{-10}}m
The diameter of the telescope is given by,
d=10cmd=10cm
The centimetre is to be converted into metres.
⇒d=10cm=10×10−2m\Rightarrow d=10cm=10\times {{10}^{-2}}m
Substituting this in the equation will give,
⇒θ=1.22×5000×10−1010×10−2=1.22×5×10−6=6×10−6rad\Rightarrow \theta =1.22\times \dfrac{5000\times {{10}^{-10}}}{10\times {{10}^{-2}}}=1.22\times 5\times {{10}^{-6}}=6\times {{10}^{-6}}rad

So, the correct answer is “Option D”.

Note:
If the distance taken is larger, the two points will be perfectly resolved. If the distance is smaller the resolution will be very poor. That is the images will not be distinctly clear. Rayleigh is the one who described this criterion using the sources having equal strength. The correction factor of 1.221.22 is obtained when the calculation regarding the position of the first circular ring of the diffraction pattern of the Airy’s disc. This circular ring is found to be dark in the central position.