A cell converts What energy

The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length

Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

$y_r=1.22\frac{\lambda f}{d}$

Where,

$y_r =$ Range of the radius

$\lambda =$ wavelength

f= focal length

Our values are given by,

State 1:

$d=2.00mm = 2*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 750nm = 750*10^{-9}m$

State 2:

$d=8.00mm = 8*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 390nm = 390*10^{-9}$

Replacing in the first equation we have:

$y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}$

$y_{r1}= 11.4\mu m$

And also for,

$y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}$

$y_{r2} = 1.49\mu m$

Therefor, the airy disk radius ranges from $1.49\mu m$ to $11.4\mu m$

7 0

3 years ago

Which states of matter have particles that move independently of one another with very little attraction

Crazy boy [7]

The state of matter that the particles move independently of one another with very little attraction is, I believe, gas

6 0

3 years ago

A long point-object, mass = 1.0 kg, moves in a circular path at a radial distance = 0.5 m from the axis of rotation. What is the

Vinvika [58]

Answer: $0.25\ kg-m^2$

Explanation:

Given

mass of Point object $m=1 kg$

Distance $r=0.5 m$

Since mass is moving in circular path therefore every time mass is at distance of r from center .

Also Moment of Inertia tells about the distribution of mass over the given region with respect to center of mass.

Therefore $I=mr^2$

$I=1\times 0.5^2$

$I=0.25\ kg-m^2$

5 0

3 years ago

A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

$q(t) = e^{-\frac{t}{(R*C)}}$