The diameter of the objective of a telescope | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The resolving power of a telescope is given by the formula:$RP = \frac{D}{1.22 \lambda}$Where $D$ is the diameter of the objective and $\lambda$ i

Vedclass · Accepted Answer

$3.28 	imes 10^{6}$

Vedclass · Answer

(D) The resolving power of a telescope is given by the formula:$RP = \frac{D}{1.22 \lambda}$Where $D$ is the diameter of the objective and $\lambda$ is the wavelength of light.Given:$D = 200 	ext{ cm} = 2 	ext{ m}$$\lambda = 5000 	ext{ \AA} = 5000 	imes 10^{-10} 	ext{ m} = 5 	imes 10^{-7} 	ext{ m}$Substituting the values:$RP = \frac{2}{1.22 	imes 5 	imes 10^{-7}}$$RP = \frac{2}{6.1 	imes 10^{-7}}$$RP = \frac{2}{6.1} 	imes 10^{7}$$RP \approx 0.32786 	imes 10^{7}$$RP \approx 3.28 	imes 10^{6}$Therefore, the resolving power of the telescope is $3.28 	imes 10^{6}$.

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 $\lambda = 5000 \text{ \AA}$.

Similar Questions

$A$ person is observing a bacteria through a compound microscope. For better analysis and to improve the resolving power,he should:

Two points separated by a distance of $0.1\,mm$ can just be resolved in a microscope when a light of wavelength $6000\ \mathring{A}$ is used. If the light of wavelength $4800\ \mathring{A}$ is used,this limit of resolution becomes.......$mm$.

If the numerical aperture $(NA)$ of a microscope is increased,then its:

Explain the resolving power of a microscope.

The headlights of a jeep are $1.2 \,m$ apart. If the pupil of the eye of an observer has a diameter of $2 \,mm$ and light of wavelength $5896 \,\text{Å}$ is used, what should be the maximum distance of the jeep from the observer if the two headlights are just separated?

Vedclass Test Series

Exam Paper Generator

Online Exam Module