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Question

(C) The limit of resolution $(	heta_R)$ for a telescope is given by the formula:$	heta_R = \frac{1.22 \lambda}{a}$Given:$\lambda = 6000 	ext{ \AA}

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$2.9 	imes 10^{-7} 	ext{ rad}$

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(C) The limit of resolution $(	heta_R)$ for a telescope is given by the formula:$	heta_R = \frac{1.22 \lambda}{a}$Given:$\lambda = 6000 	ext{ \AA} = 6000 	imes 10^{-10} 	ext{ m} = 6 	imes 10^{-7} 	ext{ m}$$a = 100 	ext{ inch} = 100 	imes 2.54 	ext{ cm} = 254 	ext{ cm} = 2.54 	ext{ m}$Substituting the values into the formula:$	heta_R = \frac{1.22 	imes 6 	imes 10^{-7}}{2.54}$$	heta_R \approx \frac{7.32 	imes 10^{-7}}{2.54} \approx 2.88 	imes 10^{-7} 	ext{ rad}$Rounding to two significant figures, we get:$	heta_R \approx 2.9 	imes 10^{-7} 	ext{ rad}$

$A$ monochromatic light of wavelength $6000 \text{ \AA}$ coming from a star is detected in a $100 \text{ inch}$ telescope. The limit of resolution of the telescope is approximately

Similar Questions

Two-point white dots are $2 \,mm$ apart on a black paper. They are viewed by an eye of pupil diameter $3 \,mm$. What is the maximum distance at which these dots can be resolved by the eye (in $\,m$)? $(\lambda = 500 \,nm)$

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.

Assertion $(A):$ To increase the resolving power of a telescope,the aperture $(a)$ of the objective lens should be large.
Reason $(R):$ The resolving power of a telescope is given by $\frac{a}{1.22 \lambda}$.

The diameter of the objective of a telescope is $200 \text{ cm}$. What is the resolving power of the telescope? Take the wavelength of light $\lambda = 5000 \text{ \AA}$.

To increase both the resolving power and magnifying power of a telescope:

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