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What is meant by resolving power of a telescope? State any one method of increasing the resolving power of an astronomical telescope.
Answer
462.9k+ views
Hint : To answer this question, we have to use the definition of the resolving power to find out the formula for the resolving power of the telescope. From that formula, we will be able to predict the method of increasing the resolving power.
Formula Used: The formula which is used in solving this question is given by
$\Rightarrow \theta = \dfrac{{1.22\lambda }}{d} $ , here $ \theta $ is the angle subtended by two objects at the telescope, $ \lambda $ is the wavelength of the light, and $ d $ is the diameter of the objective lens.
Complete step by step answer
We know that the resolving power of an optical instrument is defined as its ability to distinguish two objects which are very closely spaced, distinctly. Now, for a telescope, the angular separation between two objects is given by
$\Rightarrow \theta = \dfrac{{1.22\lambda }}{d} $ (1)
Since the resolving power is equal to the inverse of the angle subtended by the two objects on the optical instrument, so the resolving power is given as
$\Rightarrow RP = \dfrac{1}{\theta } $
From (1)
$\Rightarrow RP = \dfrac{d}{{1.22\lambda }} $
So, as can be clearly seen from the above relation that the resolving power of a telescope is directly proportional to the diameter of the objective lens, and inversely proportional to the wavelength of the light. So, one method of increasing the resolving power of a telescope can be using the objective lens of larger diameter.
Note
The resolving power, which is discussed in this question, is used in the cameras of the smartphones. Nowadays, a phone is judged on the basis of the megapixel number of the camera fitted to it. The more the mega pixel, the more quality image its camera gives. This megapixel number basically indicates the resolving power of the camera. When an image is clicked with a camera having higher megapixels, more number of details are captured in that image.
Formula Used: The formula which is used in solving this question is given by
$\Rightarrow \theta = \dfrac{{1.22\lambda }}{d} $ , here $ \theta $ is the angle subtended by two objects at the telescope, $ \lambda $ is the wavelength of the light, and $ d $ is the diameter of the objective lens.
Complete step by step answer
We know that the resolving power of an optical instrument is defined as its ability to distinguish two objects which are very closely spaced, distinctly. Now, for a telescope, the angular separation between two objects is given by
$\Rightarrow \theta = \dfrac{{1.22\lambda }}{d} $ (1)
Since the resolving power is equal to the inverse of the angle subtended by the two objects on the optical instrument, so the resolving power is given as
$\Rightarrow RP = \dfrac{1}{\theta } $
From (1)
$\Rightarrow RP = \dfrac{d}{{1.22\lambda }} $
So, as can be clearly seen from the above relation that the resolving power of a telescope is directly proportional to the diameter of the objective lens, and inversely proportional to the wavelength of the light. So, one method of increasing the resolving power of a telescope can be using the objective lens of larger diameter.
Note
The resolving power, which is discussed in this question, is used in the cameras of the smartphones. Nowadays, a phone is judged on the basis of the megapixel number of the camera fitted to it. The more the mega pixel, the more quality image its camera gives. This megapixel number basically indicates the resolving power of the camera. When an image is clicked with a camera having higher megapixels, more number of details are captured in that image.
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