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Question: The resolving power of telescope of aperture 100cm for light of wavelength \[5.5 \times {10^{ - 7}}m...

The resolving power of telescope of aperture 100cm for light of wavelength 5.5×10−7m5.5 \times {10^{ - 7}}mis
A. 0.149×1070.149 \times {10^7}
B. 1.49×1071.49 \times {10^7}
C. 14.9×10714.9 \times {10^7}
D. 149×107149 \times {10^7}

Explanation

Solution

The resolving power is one of the special features of the telescope. It is the ability of the telescope to distinguish clearly between two points where the angular separation between them is less than the smallest angle that the observer’s eye can resolve.In this question the wavelength of the light is given and the diameter is also given so by using the resolving power formula we will find the resolving power of the telescope.

Formula Used:
RP=1dθRP = \dfrac{1}{{d\theta }}
where, dθ=1.22λDd\theta = \dfrac{{1.22\lambda }}{D}
Where, dθd\theta is the angle subtended by two distant objects at objective, λ\lambda is the wavelength of the light and D is the diameter of the telescope

Complete step by step answer:
The aperture which is the diameter of telescope D=100cm=1mD = 100cm = 1m
The wavelength of the light λ=5.5×10−7m\lambda = 5.5 \times {10^{ - 7}}m
Now since the wavelength of the light and the diameter of the telescope is given so we will find the angle subtended by two distant objects given by the formula
θ=1.22λD\theta = \dfrac{{1.22\lambda }}{D}
Hence by substituting the values in the equation we get

dθ=1.22×5.5×10−71 ⇒dθ=6.71×10−7 d\theta = \dfrac{{1.22 \times 5.5 \times {{10}^{ - 7}}}}{1} \\\ \Rightarrow d\theta = 6.71 \times {10^{ - 7}} \\\

So the resolving power of the telescope will be,
RP=1dθ=16.71×10−7=0.149×107RP = \dfrac{1}{{d\theta }} = \dfrac{1}{{6.71 \times {{10}^{ - 7}}}} = 0.149 \times {10^7}
Therefore, the resolving power of telescope of aperture 100cm100cm for light of wavelength 5.5×10−7m5.5 \times {10^{ - 7}}m will be0.149×107 0.149 \times {10^7}

Hence,option A is correct.

Note: The resolution of a telescope depends on the nature of the light it is dealing with, it depends on the wavelength of the light which is incoming light in the telescope. Students can increase or decrease the resolving power of the telescope by varying the diameter of the objective lens of the telescope and the resolving power is directly proportional to the diameter of the lens.