If an object kept at the least distance of di | vedclass

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

Question

(A) The angular resolution of the eye is given by $	heta = 1.22 \frac{\lambda}{D}$,where $\lambda$ is the wavelength and $D$ is the diameter of the p

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The angular resolution of the eye is given by $	heta = 1.22 \frac{\lambda}{D}$,where $\lambda$ is the wavelength and $D$ is the diameter of the pupil.For an object at distance $x$ and size $h$,the angle subtended is $	heta = \frac{h}{x}$.Since the object is 'just resolved',we have $\frac{h}{x} \propto \frac{\lambda}{D}$.Thus,$x \propto \frac{hD}{\lambda}$. Assuming the object size $h$ remains constant,we have $\frac{x_2}{x_1} = \frac{D_2}{D_1} 	imes \frac{\lambda_1}{\lambda_2}$.Given $x_1 = 25 \ cm$,$\lambda_1 = 500 \ nm$,$D_1 = 1 \ mm$,$\lambda_2 = 400 \ nm$,and $D_2 = 0.8 \ mm$.Substituting the values: $x_2 = 25 	imes \frac{0.8}{1} 	imes \frac{500}{400} = 25 	imes 0.8 	imes 1.25 = 25 	imes 1 = 25 \ cm$.

If an object kept at the least distance of distinct vision is just resolved with light of wavelength $500 \ nm$ and a pupil of diameter $1 \ mm$,at what distance will the object be just resolved if the wavelength is $400 \ nm$ and the pupil diameter is $0.8 \ mm$ (in $cm$)?

Similar Questions

$A$ microscope has an objective of aperture $8 \text{ mm}$ and focal length of $5 \text{ cm}$. The minimum separation between two objects to be just resolved by the microscope is (wavelength of light used $= 5500 \text{ Å}$) (in $\mu\text{m}$)

The human eye has an approximate angular resolution of $\phi = 5.8 \times 10^{-4} \, rad$ and a typical photoprinter prints a minimum of $300 \, dpi$ (dots per inch, $1 \, inch = 2.54 \, cm$). At what minimal distance $z$ should a printed page be held so that one does not see the individual dots?

The diameter of the objective lens of a telescope is $250\, cm$. For light of wavelength $600\, nm$ coming from a distant object, the limit of resolution of the telescope is close to:

The resolving limit of a healthy human eye is about:

The diameter of the objective of a telescope is $a$,its magnifying power is $m$,and the wavelength of light is $\lambda$. The resolving power of the telescope is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module