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(D) The angular resolution $	heta$ of a telescope is given by the formula $	heta = 1.22 \frac{\lambda}{a}$, where $\lambda$ is the wavelength of lig

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$30$

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(D) The angular resolution $	heta$ of a telescope is given by the formula $	heta = 1.22 \frac{\lambda}{a}$, where $\lambda$ is the wavelength of light and $a$ is the diameter of the objective lens.Given: $	heta = 3.0 	imes 10^{-7}$ radian, $\lambda = 500 	ext{ nm} = 500 	imes 10^{-9} 	ext{ m}$.Rearranging the formula to solve for $a$: $a = \frac{1.22 \lambda}{	heta}$.Substituting the values: $a = \frac{1.22 	imes 500 	imes 10^{-9}}{3.0 	imes 10^{-7}} = \frac{610 	imes 10^{-9}}{3.0 	imes 10^{-7}} = 203.33 	imes 10^{-2} 	ext{ m} = 2.0333 	ext{ m}$.Converting to centimeters: $2.0333 	ext{ m} = 203.33 	ext{ cm}$.The nearest integer is $203 	ext{ cm}$. Note: Given the provided options, there appears to be a discrepancy in the question's expected range; however, based on the physics formula, the calculated value is $203 	ext{ cm}$.

Similar Questions

Assuming the human pupil to have a radius of $0.25 \ cm$ and a comfortable viewing distance of $25 \ cm$,the minimum separation between two objects that the human eye can resolve at $500 \ nm$ wavelength is..... $\mu m$.

Calculate the limit of resolution of a telescope objective having a diameter of $200\, cm$, if it has to detect light of wavelength $500\, nm$ coming from a star.

If the wavelengths of light used in an optical instrument are $\lambda_1 = 4000 \, \mathring A$ and $\lambda_2 = 5000 \, \mathring A$,what will be the ratio of their resolving powers?

Assertion: The resolving power of a telescope is more if the diameter of the objective lens is more.
Reason: Objective lens of large diameter collects more light.

When the object is self-luminous,the resolving power of a microscope is given by the expression

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