Resolving power-Human eye, Microscopes and Telescopes Questions in English

(D) The distance between the two point dots is $x = 2 \,mm = 2 \times 10^{-3} \,m$.
The diameter of the pupil is $d = 3 \,mm = 3 \times 10^{-3} \,m$.
The wavelength of light is $\lambda = 500 \,nm = 5 \times 10^{-7} \,m$.
According to the Rayleigh criterion for the resolution of two points, the angular separation $\theta$ is given by $\theta = \frac{1.22 \lambda}{d}$.
Also, the angular separation can be expressed as $\theta = \frac{x}{D}$, where $D$ is the maximum distance.
Equating the two expressions: $\frac{x}{D} = \frac{1.22 \lambda}{d}$.
Rearranging for $D$: $D = \frac{x d}{1.22 \lambda}$.
Substituting the values: $D = \frac{(2 \times 10^{-3} \,m) \times (3 \times 10^{-3} \,m)}{1.22 \times (5 \times 10^{-7} \,m)}$.
$D = \frac{6 \times 10^{-6}}{6.1 \times 10^{-7}} = \frac{60}{6.1} \approx 9.836 \,m$.
Rounding to the nearest integer, we get $D \approx 10 \,m$.

(A) The resolving power of a microscope is given by $R = \frac{1}{d} = \frac{2n \sin \beta}{1.22 \lambda}$,where $d$ is the minimum separation between two objects.
Thus,the minimum separation $d$ is inversely proportional to the refractive index $n$ of the medium $(d \propto \frac{1}{n})$.
Given for air $(n_1 = 1)$,$d_1 = 6 \mu m$.
For a medium with refractive index $n_2 = 1.5$,the new minimum separation $d_2$ is calculated as:
$d_2 = \frac{d_1 \times n_1}{n_2} = \frac{6 \mu m \times 1}{1.5} = 4 \mu m$.
Therefore,the minimum separation is $4 \mu m$.

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

(B) The formula for the limit of resolution of a microscope is given by $d = \frac{\lambda}{2 NA}$.
Given:
Numerical aperture,$NA = 0.8$
Wavelength,$\lambda = 0.6 \mu m$
Substituting these values into the formula:
$d = \frac{0.6}{2 \times 0.8}$
$d = \frac{0.6}{1.6}$
$d = \frac{6}{16} \mu m$
$d = \frac{3}{8} \mu m$
Therefore,the correct option is $B$.

(C) The minimum separation $(d_{\min})$ between two objects that can be just resolved by a microscope is given by the formula:
$d_{\min} = \frac{1.22 f \lambda}{D}$
Given:
$f = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$
$\lambda = 5500 \text{ Å} = 5500 \times 10^{-10} \text{ m}$
$D = 8 \text{ mm} = 8 \times 10^{-3} \text{ m}$
Substituting the values:
$d_{\min} = \frac{1.22 \times 5 \times 10^{-2} \times 5500 \times 10^{-10}}{8 \times 10^{-3}}$
$d_{\min} = \frac{1.22 \times 5 \times 5500 \times 10^{-12}}{8 \times 10^{-3}}$
$d_{\min} = \frac{33550 \times 10^{-12}}{8 \times 10^{-3}}$
$d_{\min} = 4193.75 \times 10^{-9} \text{ m} \approx 4.19 \times 10^{-6} \text{ m}$
$d_{\min} \approx 4.2 \mu\text{m}$

(A) Given: Diameter of pupil, $d = 2 \,mm = 2 \times 10^{-3} \,m$. Distance of objects, $D = 20 \,m$. Wavelength of light, $\lambda = 600 \,nm = 6 \times 10^{-7} \,m$.
The limit of resolution for a circular aperture (like the pupil of an eye) is given by the Rayleigh criterion:
$\theta = \frac{1.22 \lambda}{d}$
The minimum separation $y$ between two objects at distance $D$ is given by $y = \theta \times D$.
Substituting the values:
$y = \frac{1.22 \times 6 \times 10^{-7} \times 20}{2 \times 10^{-3}}$
$y = 1.22 \times 6 \times 10^{-4} \times 10 = 7.32 \times 10^{-3} \,m$
$y = 7.32 \,mm$.
Thus, the minimum separation the human eye can resolve is $7.32 \,mm$.

(B) Given:
Diameter of the objective,$d = 3.6 \ m$
Wavelength of light,$\lambda = 540 \ nm = 540 \times 10^{-9} \ m$
The limit of resolution $(d\theta)$ of a telescope is given by the formula:
$d\theta = \frac{1.22 \lambda}{d}$
Substituting the values:
$d\theta = \frac{1.22 \times 540 \times 10^{-9} \ m}{3.6 \ m}$
$d\theta = \frac{658.8 \times 10^{-9}}{3.6} \ rad$
$d\theta = 183 \times 10^{-9} \ rad$
$d\theta = 1.83 \times 10^{-7} \ rad$

(A) The limit of resolution or angular resolution for a telescope is given by the formula: $\alpha_{\min} = \frac{1.22 \lambda}{D}$.
Given values are:
$\alpha_{\min} = 3.0 \times 10^{-7} \text{ rad}$
$\lambda = 525 \text{ nm} = 525 \times 10^{-9} \text{ m}$
Rearranging the formula to solve for the diameter of the objective lens $(D)$:
$D = \frac{1.22 \lambda}{\alpha_{\min}}$
Substituting the values:
$D = \frac{1.22 \times 525 \times 10^{-9}}{3.0 \times 10^{-7}}$
$D = \frac{640.5 \times 10^{-9}}{3.0 \times 10^{-7}}$
$D = 213.5 \times 10^{-2} \text{ m} = 2.135 \text{ m}$.
Rounding to the nearest given option, we get $D = 2.1 \text{ m}$.

(A) The limit of resolution of a telescope is given by the formula: $\alpha = \frac{1.22 \lambda}{a}$, where $\alpha$ is the angular resolution, $\lambda$ is the wavelength of light, and $a$ is the diameter of the objective lens.
Given values are $\alpha = 2.5 \times 10^{-7} \text{ rad}$ and $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}$.
Rearranging the formula for $a$: $a = \frac{1.22 \lambda}{\alpha}$.
Substituting the values: $a = \frac{1.22 \times 500 \times 10^{-9}}{2.5 \times 10^{-7}}$.
$a = \frac{610 \times 10^{-9}}{2.5 \times 10^{-7}} = 244 \times 10^{-2} \text{ m} = 2.44 \text{ m}$.
Converting to centimeters: $2.44 \text{ m} = 244 \text{ cm}$.
Thus, the diameter of the objective lens is $244 \text{ cm}$.

(A) The limit of resolution of a telescope is given by the formula:
$\theta = \frac{1.22 \lambda}{d}$
where $\lambda$ is the wavelength of light and $d$ is the diameter of the objective lens.
Given:
$\lambda = 6400 \text{ Å} = 6400 \times 10^{-10} \,m = 6.4 \times 10^{-7} \,m$
$d = 200 \,cm = 2 \,m$
Substituting these values into the formula:
$\theta = \frac{1.22 \times 6.4 \times 10^{-7}}{2}$
$\theta = 1.22 \times 3.2 \times 10^{-7}$
$\theta = 3.904 \times 10^{-7} \,rad$
Rounding to the nearest significant value, we get $\theta \approx 3.9 \times 10^{-7} \,rad$.

(A) The resolving limit $(d\theta)$ of a telescope is given by the formula:
$d\theta = \frac{1.22 \lambda}{a}$
where $\lambda$ is the wavelength of light and $a$ is the diameter of the objective lens.
Given:
$\lambda = 4538 \ \text{Å} = 4538 \times 10^{-10} \ \text{m}$
$a = 1 \ \text{m}$
Substituting the values:
$d\theta = \frac{1.22 \times 4538 \times 10^{-10}}{1}$
$d\theta = 5536.36 \times 10^{-10} \ \text{rad}$
$d\theta \approx 5.54 \times 10^{-7} \ \text{rad}$

(C) The angular resolution $\theta$ is given by the ratio of the distance between two dots $x$ to the distance $d$ from the eye, i.e., $\theta = \frac{x}{d}$.
Given that the printer prints $300 \text{ dots per inch}$, the distance between two adjacent dots $x$ is $x = \frac{1 \text{ inch}}{300} = \frac{2.54 \text{ cm}}{300}$.
Substituting the values into the formula: $5.8 \times 10^{-4} = \frac{2.54 \text{ cm} / 300}{d}$.
Rearranging for $d$: $d = \frac{2.54}{300 \times 5.8 \times 10^{-4}} \text{ cm}$.
$d = \frac{2.54}{0.174} \text{ cm} \approx 14.597 \text{ cm}$.
Thus, the minimal distance is approximately $14.59 \text{ cm}$.

(B) The angular resolution of a telescope is given by the formula $\theta = 1.22 \frac{\lambda}{R}$.
Given:
$\theta = 5 \times 10^{-7} \text{ rad}$
$\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}$
Substituting these values into the formula:
$5 \times 10^{-7} = 1.22 \times \frac{500 \times 10^{-9}}{R}$
Rearranging for $R$:
$R = \frac{1.22 \times 500 \times 10^{-9}}{5 \times 10^{-7}}$
$R = 1.22 \times 100 \times 10^{-2}$
$R = 1.22 \text{ m}$
Converting to centimeters:
$R = 1.22 \times 100 \text{ cm} = 122 \text{ cm}$.

(D) The angular resolution $\theta$ of a telescope is given by the formula $\theta = 1.22 \frac{\lambda}{a}$,where $\lambda$ is the wavelength of light and $a$ is the diameter of the objective lens.
Given: $\theta = 3.0 \times 10^{-7}$ radian,$\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}$.
Rearranging the formula to solve for $a$: $a = \frac{1.22 \lambda}{\theta}$.
Substituting the values: $a = \frac{1.22 \times 500 \times 10^{-9}}{3.0 \times 10^{-7}} = \frac{610 \times 10^{-9}}{3.0 \times 10^{-7}} = 203.33 \times 10^{-2} \text{ m} = 2.0333 \text{ m}$.
Converting to centimeters: $2.0333 \text{ m} = 203.33 \text{ cm}$.
The nearest integer is $203 \text{ cm}$. 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 $203 \text{ cm}$.