Determine the minimum number of photons that must enter the eye's pupil each second in order for an object to be seen. Assume that the pupil's radius is 0.20 cm and the wavelength of the light is 550 nm . To see an object with the unaided eye, the light intensity coming to the eye must be about 5×10−12J/m2⋅s or greater. Calculate Nmin... I keep getting 1.38 e25 which is not right and I can figure out what I am doing wrong.

Explanation:

Energy of a photon = hν , h is plank's constant and ν is frequency .

=    h c / λ     ; c is velocity of light and  λ is wavelength .

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 550 x 10⁻⁹

= .036 x 10⁻¹⁷ J  

Let no of photons be n .

Total energy of n photons

= n x  .036 x 10⁻¹⁷ J  

Intensity of light = 5 x 10⁻¹² J /m²

Total energy flowing through eye per second

= intensity x area of eye.

=   5 x 10⁻¹² x π x .2² x 10⁻⁴

= .628 x 10⁻¹⁶ J

so

n x  .036 x 10⁻¹⁷  =  .628 x 10⁻¹⁶

n = .628 x 10⁻¹⁶ /  .036 x 10⁻¹⁷

= 17.36 x 10

= 173.6

174.

convex lenses are thicker at the middle. rays of light that pass through the lens are brought closer together (they converge). a convex lens is a converging lens. when parallel rays of light pass through a convex lens the refracted rays converge at one point called the principal focus.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.