Minimum light intensity that can be perceived by normal human eye is about `10^(-10) Wm^(-2)`. What is the minimum number of photons of wavelength 660 nm that must enter the pupil in one second, for one to see the object?Area of cross-section of the pupil is `10^(-4)m^(2)`?

The minimum intensity of light to be detected by human eye is 10^(-10) W//m^(2) . The number of photons of wavelength 5.6 xx 10^(-7) m entering the eye , with pupil area 10^(-6) m^(2) , per second for vision will be nearly

The minimum intensity of light to be detected by human eye is 10^(-10) W//m^(2) . The number of photons of wavelength 5.6 xx 10^(-7) m entering the eye , with pupil area 10^(-6) m^(2) , per second for vision will be nearly

The pupil allows light to pass through it and has a diameter of 2xx10^(-3)m. If minimum light intensity that should enter the eye is 1.8xx10^(-10) W//m^(-2) , then find the number of photon in visible range must enter the pupil to have vision. Wavelength of photon in visible range can be taken as 5,500Å .

A star which can be seen with naked eye fro Earth has intensity 1.6xx10^(-9)Wm^(-2) on Earth.If the corresponding wavelength is 560 nm,and the radius of the lens of human eye is 2.5xx10^(-3) m,the number of photons entering in our eye in 1s is…….